The elements of abstract algebra may look like a heap of definitions, and when it starts to become really interesting it also gets quite difficult. This page is just a start. Nevertheless I hope you will understand how earlier known results get a common ground. This page will also provide some necessary background for coming surveys on Galois Theory and Algebraic Number Theory. My intention is to make at least some essential part of these important and interesting areas of mathematics understandable to interested students at senior high school level or slightly above.
In examples we will use the standard notations Z, Q, R, C for the set of integers, rational numbers, real numbers, and complex numbers, respectively.
A binary operation on a nonempty set M is a function from M×M to M. 
A binary operation can be defined by means of a table if the set is
finite. An example where the set is {a,b,c,d}:
*  a  b  c  d 

a  a  c  d  b 
b  c  d  a  b 
c  d  a  b  c 
d  b  b  c  a 
A nonempty set together with one or several binary operations is called
an algebraic structure. An algebraic structure with set M
and operation * is denoted (M,*) or just M. Instead
of * the usual multiplication dot is common (although maybe not with its
usual meaning). As usual the dot can often be left out. A commutative operation
is often denoted + and if there are two operations defined, one of them
commutative, + and · are common. Then the structure may be denoted
(R,+,·) . We will now have a look at three basic structures,
groups, rings, and fields.
A group is an algebraic structure (G,*) such that
(ii) there is a neutral element e in G such that a*e = e*a = a for all a in G, (iii) for each a in G there is an inverse a' in G such that a*a' = a'*a = e. 
Example 2: No infinite subset of Z is a group under multiplication, but ({1,1},·) is a group.
Example 3: Let Z_{3}
denote the algebraic structure with elements 0,1,2 and the operation +
defined by the group table
+  0  1  2 

0  0  1  2 
1  1  2  0 
2  2  0  1 
In a similar way a group (Z_{n},+)
can be defined for any positive integer n. However, generalizing
(Z_{3}
 {0},·) to (Z_{n}
 {0},·) results in a group if and only if n is a prime
number.
·  e  a  b  c  ·  e  a  b  c  

e  e  a  b  c  e  e  a  b  c  
a  a  e  c  b  a  a  e  c  b  
b  b  c  e  a  b  b  c  a  e  
c  c  b  a  e  c  c  b  e  a 
We have found that there are exactly two nonisomorphic groups of order 4, both abelian. The first one is called Klein's Fourgroup while the second one is isomorphic to (Z_{4},+).
Up to now all groups have been abelian. The smallest nonabelian group is of order 6. If the elements are {e,p,q,r,s,t} its group table can be written:
*  e  p  q  r  s  t 

e  e  p  q  r  s  t 
p  p  q  e  t  r  s 
q  q  e  p  s  t  r 
r  r  s  t  e  p  q 
s  s  t  r  q  e  p 
t  t  r  s  p  q  e 
Permutation groups played a central role in the development of group theory and are essential in Galois theory. The symmetric group S_{n} is defined as the permutation group of n elements, and has n! elements. The even permutations make up a subgroup, the alternating group A_{n}, with n!/2 elements. Also more general groups of functions will appear. Notice that the order in which functions are applied is in general essential so such groups are normally nonabelian.
A subgroup is defined as a group, whose elements form a subset
of a group, with the same group operation. There are always at least two
subgroups, one being the entire group, the other consisting of just the
neutral element. These are called trivial subgroups. By a common
abuse of notation, the latter is often denoted 0 or 1. All other subgroups
are called proper subgroups. A nonempty subset H of a group
G is a subgroup of G if and only if ab^{1}
is in H whenever a and b are in H. Here b^{1}
denotes the inverse (in G) of b. We prove this:
Conversely, suppose H is a subgroup of G. Then there is
a neutral element e in H, and any b in H has
an inverse b^{1} in H. Since
H is a subgroup, it is closed under the group operation of G,
so ab^{1} is in H if a
and b are in H, since b^{1}
is then in H. Notice, that e and b^{1}
are neutral element and inverse, respectively, in G also, since
the group operation is the same.
Now, let a be any element of a group G. Then all powers a^{n} (n a nonnegative integer) will make up an abelian subgroup of G. Such a group is said to be generated by a, and is called a cyclic group. If G has prime order and a is not the neutral element then the generated subgroup must be the entire group. So, a group of prime order is cyclic and abelian.
Finally, let us consider the group of symmetry operations of the square. A symmetry operation can be described as a displacement of the square such that it occupies exactly the same place after as before being displaced. The operations are definied by the final position relative to the original only, not how the motion is performed. There are three different clockwise rotations of 90, 180, and 270 degrees, respectively. Notice that a counterclockwise rotation equals some clockwise rotation giving the same resulting position. Further, there are reflections in the diagonals and in the horisontal and vertical lines bisecting the square. No displacement at all is also a symmetry operation, making 8 in all. A symmetry operation can be regarded as a function, taking one position of the square as its input and the resulting relative position as its value. It is then obvious that the symmetry operations constitute a group, the symmetry group of the square, denoted D_{8}. The group operation is composition of functions.
If the corners are numbered as in the above picture, each symmetry operation corresponds to a permutation of 1,2,3,4. However, not all permutation occur since the ordering of the corners cannot be changed. Thus, the symmetry group must be (isomorphic to) a proper subgroup of S_{4}. Since S_{4} has 4! = 24 elements this is in agreement with Lagrange's theorem.
Similarly, there are symmetry groups for other figures. These groups play a role in Galois theory. For example, the so called Galois Group of the polynomial t^{4}  2 over Q is isomorphic to D_{8}.
We will return to permutation groups and subgroups later, in particular so called normal subgroups.
A ring is an algebraic structure (R,+,·) such
that
(ii) · is associative, (iii) a·(b + c) = a·b + a·c and (a + b)·c = a·c + b·c for all a, b, c in R. 
A field is a ring (K,+,·) such that

Example 2: (Z_{n},+,·) is a ring for any positive integer n. It is a field if and only if n is a prime number.
Example 3: In the same way as for groups, the operations + and · can be defined by tables. The following tables define the operations for (Z_{4},+,·), which is a ring but not a field:
+  0  1  2  3  ·  0  1  2  3  

0  0  1  2  3  0  0  0  0  0  
1  1  2  3  0  1  0  1  2  3  
2  2  3  0  1  2  0  2  0  2  
3  3  0  1  2  3  0  3  2  1 
Example 1: (Z,+,·) is a domain but not a field.
Example 2: (Z_{n},+,·) has zerodivisors unless n is a prime number.
Polynomials play an important role in Galois theory and in algebraic
number theory. A polynomial in one indeterminate t with coefficients
in a ring R is an expression of the form
Part of the theory of factorization is taken over from real polynomials and part is new. For example any polynomial in K[t] can, apart from a constant factor, be uniquely factorized into irreducible polynomials. An irreducible polynomial is one which cannot be written as a product of nonconstant polynomials of lower degree.
Example: In Z_{3}[t] , 2t^{2} + 2t + 2 = (t + 2)(2t + 1) = 2(t + 2)(t + 2) , since 2·2 = 1 in Z_{3}.
A subring is defined as a ring, whose elements form a subset
of a ring, with the same binary operations. There are always at least two
subrings, the entire ring and {0}. These are called trivial subrings.
By a common abuse of notation, the latter is often denoted 0. All other
subrings are called proper subrings. A nonempty subset S
of a ring R is a subring of R if and only if a 
b and ab are in S whenever a and b
are in S and 1 (of R) is in S. This is proved similarly
to the corresponding theorem for subgroups.
A group homomorphism f is a function from a group G
to a group G' such that

A ring homomorphism f is a function from a ring R
to a ring R' such that
(ii) f(a·b) = f(a)·f(b) for all a, b in R. 
For a group homomorphism f(e) is the neutral element of
G' and f(a^{1})
= (f(a))^{1} . We prove
this:
If two finite groups or rings are isomorphic the tables defining their operations are the same except for possible renaming and rearrangement of the elements. In this sense they are essentially the same, and often their difference does not matter. The same goes for infinite groups and rings although the tables cannot actually be completely written out.
Example 1: The complex numbers with imaginary part 0 are usually thought of as R. More precisely, they form a subfield of C, isomorphic to R. This distinction does not matter, e g when solving polynomial equations. On the contrary it is easiest to identify the subfield with R.
Example 2: For any complex number z = a + bi let f(z) = a  bi . Then f is an automorphism of C. Since f(z) = z when z is real, but not else, f is called an Rautomorphism of C.
Example 3: Let f be as in example 2, and let g be the identity function on C. Then g is, of course, also an Rautomorphism of C. f and g are, in fact, the only Rautomorphisms of C. The constitute a group with composition of functions as group operation, the so called Galois Group of the field extension C:R or of the polynomial t^{2} + 1 over R. This group is isomorphic to (Z_{2},+) and to S_{2}.
Example 4: In connection with the symmetry operations of a square, we noted that these operations can be put into a 11 correspondence with eight of the permutations of 1,2,3,4. Now, let f(x) for the symmetry operation x equal the corresponding permutation of 1,2,3,4. Then f is a homomorphism from D_{8} to S_{4}. It is injective but not surjective, so it is a monomorphism but no isomorphism. Thus D_{8} is isomorphic to a proper subgroup of S_{4}.
For future reference we define:
(iii) If f is a group or ring homomorphism the image im(f) is the set of values taken on by f.
H is a normal subgroup of the group G if
ghg^{1} is in H
for all g in G and h in H. 
Example 1: The trivial subgroups are always normal.
Example 2: ({e,p,q},*) is a normal subgroup of ({e,p,q,r,s,t},*) .
Example 3: If f is a group homomorphism from G
to G' then ker(f) is a normal subgroup of G.
We prove this:
If H is a normal subgroup of the group G, then
the set {gH: g in G}
with a binary operation · defined by

Since H is normal aH = Ha, and bH
= Hb. From this a_{1}b_{1}H
= a_{1}Hb_{1}
= a_{2}Hb_{1}
= a_{2}b_{1}H
= a_{2}b_{2}H
if
a_{1}H = a_{2}H,
and b_{1}H = b_{2}H.
eH = H is obviously neutral element, and a^{1}H inverse of aH, as usual unique.
Finally (aHbH)cH = (abH)cH = (ab)cH
= a(bc)H = aH(bcH), so · is associative.
·  H  rH 

H  H  rH 
rH  rH  H 
Example 2: Let f be a group homomorphism from G to G'. Then there is always a quotient group G/ker(f) since the kernel ker(f) is a normal subgroup of G.
In example 2 we noticed the existence of the group G/ker(f).
If, in particular, f is injective, i e a monomorphism, then G
is isomorphic to im(f), and ker(f) = ({e},*).
We prove the latter statement:
If f is a group homomorphism from G to G', then
G/ker(f) is isomorphic to im(f). 
F is obviously surjective, since f(x) takes all values of im(f) when x goes through all values of G.
Next, let F(xker(f)) = F(xker(f)), i e f(x) = f(y). Then, since f is a homomorphism, f(xy^{1}) = f(x)f(y^{1}) = f(x)(f(y))^{1} = f(x)(f(x))^{1} = e', so xy^{1} is in ker(f). But then xker(f) = yker(f), i e the same element in G/ker(f), so F is injective.
Finally, F is also a homomorphism because F(xker(f)·yker(f))
= F(xyker(f)) = f(xy) = f(x)f(y)
= F(xker(f))F(yker(f)) .
Example 2: Let G be a finite group with a normal subgroup H. Then H and G/H are also finite. We use   to denote the number of elements in a finite set. From Lagrange's theorem, G/H is an integer. Now, all cosets aH have the same number of elements, equal to H. Thus, the number of cosets must be G/H, but it is also G/H, so G/H = G/H. Further, let f be a homomorphism defined on G. Then im(f) and G/ker(f) have the same number of elements, since they are isomorphic. From this we deduce G = ker(f)·im(f).
The ideals have their roots in the 19th century attempt to restore unique prime factorization in so rings of algebraic integer.
A nonempty subset I of a ring R is an ideal in
the ring R if
(ii) rx is in I for all r in R and x in I. 
Example 1: In any ring R there are always at least the ideals {0} and R.
Example 2: For any natural number n, let nZ = {nx:x in Z}. The nZ are ideals of Z, in fact the only ones. For positive n, nZ can be described as the set of integers divisible by n. In particular 2Z is the set of even numbers.
Example 3: If f is a ring homomorphism from R
to R' then ker(f) is an ideal of R. We
prove this:
If I is an ideal in the ring R, then the set {r
+ I: r in R} with binary operations + and ·
defined by
(ii) (r + I)·(s + I) = (rs) + I 
For + this is in principle as for quotient groups since (R,+)
is an abelian group, and thus (I,+) a normal subgroup.
For · suppose r_{1}
+ I = r_{2} + I
and s_{1} + I = s_{2}
+ I, i e r_{1}  r_{2}
and s_{1}  s_{2}
are in I, or r_{1} =
r_{2} + i_{1}
and s_{1} = s_{2}
+ i_{2} where i_{1},
i_{2} are in I. We must show
that (r_{1}s_{1})
+ I = (r_{2}s_{2})
+ I. This is true if and only if r_{1}s_{1}
 r_{2}s_{2}
is in I, which holds since r_{1}s_{1}
 r_{2}s_{2}
= (r_{2} + i_{1})(s_{2}
+ i_{2})  r_{2}s_{2}
= r_{2}i_{2}
+ s_{2}i_{1}
+ i_{1}i_{2},
where every term has a factor in I.
0 + I = I is obviously neutral element for +, and (r) + I inverse of r + I, as usual unique.
Since ((r + I) + (s + I)) + (t + I) = ((r + s) + I) + (t + I) = ((r + s) + t) + I = (r + (s + t)) + I = (r + I) + ((s + t) + I) = (r + I) + ((s + I) + (t + I)), so + is associative. It is also commutative since + in R is. Thus, (R/I,+) is an abelian group.
Similarly ((r + I)·(s + I))·(t + I) = (r + I)·((s + I)·(t + I)), so · is associative.
Finally (r + I)·((s + I) +
(t + I)) = (r + I)·((s + t)
+ I) = r(s + t) + I = (rs + rt)
+ I = ((rs) + I) + ((rt) + I) = (r
+ I)·(s + I) + (r + I)·(t
+ I), which means that the distributive laws hold.
+  4Z  1+4Z  2+4Z  3+4Z  ·  4Z  1+4Z  2+4Z  3+4Z  

4Z  4Z  1+4Z  2+4Z  3+4Z  4Z  4Z  4Z  4Z  4Z  
1+4Z  1+4Z  2+4Z  3+4Z  4Z  1+4Z  4Z  1+4Z  2+4Z  3+4Z  
2+4Z  2+4Z  3+4Z  4Z  1+4Z  2+4Z  4Z  2+4Z  4Z  2+4Z  
3+4Z  3+4Z  4Z  1+4Z  2+4Z  3+4Z  4Z  3+4Z  2+4Z  1+4Z 
Example 3: Let f be a ring homomorphism from R
to R'. Then there is always a quotient ring R/ker(f)
since the kernel ker(f) is an ideal in R.
If f is a ring homomorphism from R to R', then
R/ker(f) is isomorphic to im(f). 
F is obviously surjective, since f(x) takes all values of im(f) when x goes through all values of R.
Next, let F(x + ker(f)) = F(x + ker(f)), i e f(x) = f(y). Then, since f is a homomorphism, f(x  y) = f(x)  f(y) = f(x)  f(x) = 0, so x  y is in ker(f). But then x + ker(f) = y + ker(f), i e the same element in R/ker(f), so F is injective.
Finally, F is also a homomorphism because F((x
+ ker(f)) + (y + ker(f))) = F((x
+ y) + ker(f)) = f(x + y) = f(x)
+ f(y) = F(x + ker(f)) + F(y
+ ker(f)), F((x + ker(f))·(y
+ ker(f))) = F((xy) + ker(f)) = f(xy)
= f(x)f(y) = F(x + ker(f))·F(y
+ ker(f)), and F(1 + ker(f)) =
f(1) = 1'.
Example 2: Let I, J be ideals in the ring R,
and J contained in I. Then there is a homomorphism f
from R/J onto R/I, defined by f(x
+ J) = x + I. f(x + J)
= I, the zero element of R/I, if and only if x
is in I, so ker(f) = I/J. Thus, (R/J)/(I/J)
is isomorphic to R/I.
In most cases a field extension can be considered as an ordered pair of two fields K and L, where K is a subfield of L. The extension is then denoted L:K.The precise definition says that a field extension is a monomorphism from a field K to a field L. However, usually isomorphic fields can be identified, so that K is a subfield of L. We also say that L is an extension of K. K and L are called the small field and the large field, respectively.
Example 1: R:Q and C:Q are field extensions.
Example 2: Let Q(sqrt(2)) denote the smallest field containg Q and sqrt(2). Being a field, Q(sqrt(2)) must contain elements such as 1/(1 + sqrt(2)) = sqrt(2)  1. It is not difficult to prove that all elements of Q(sqrt(2)) can be written a + bsqrt(2) with rational numbers a, b.
Example 3: Let <t^{2}
+ 1> = (t^{2} + 1)R[t].
Then <t^{2} + 1> is an ideal in
the ring R[t], and R[t]/<t^{2}
+ 1> is a ring. This ring is isomorphic to C,
so it is a field. We prove this:
We must also show that im(F) = C. To this end let z be any complex number and f(t) = Re z + tIm z. Then f(t) is a polynomial in R[t] with f(i) = z. So, for any z in C there is a f(t) in R[t] such that F(f(t)) = f(i) = z. This means that F is surjective, i e im(F) = C. Since ker(F) = <t^{2} + 1>, it follows that R[t]/<t^{2} + 1> is isomorphic to C.
Definition:
A simple algebraic extension is an extension K(a):K
such that a is a zero
of a nonzero polynomial over K. a is then said to be algebraic over K. 
A simple transcendental extension is an extension K(a):K
where a is not
a zero of any nonzero polynomial over K. a is then said to be transcendental over K. 
Example 2: R[t]/<t^{2} + 1> is a simple algebraic extension of R. In this case R is not a subfield, but R is isomorphic to a subfield, namely {a + <t^{2} + 1>:a in R}.
Complex numbers, which are algebraic over Q, are simply called algebraic numbers. They constitute a field, and are studied in algebraic number theory. Complex numbers, which are not algebraic, are called transcendental. Some of the more famous transcendental numbers are e, pi, and 2^{sqrt(2)}.
For the next theorem we notice that a polynomial over a field, with highest degree coefficient 1, is said to be monic.
If K(a):K is a simple algebraic extension, then
there is a unique,
monic polynomial m(t) over K of lowest degree, such that m(a) = 0. The polynomial m(t) is irreducible, and divides every polynomial over K, of which a is a zero. 
Next, suppose m(t) = f(t)g(t). Then f(a)g(a) = 0, so f(a) = 0 or g(a) = 0, since f(a) and g(a) are elements of the field K. This implies that a is a zero of a polynomial of lower degree than m(t), which means a contradiction. Thus, m(t) is irreducible.
Finally, suppose a is a zero of a polynomial f(t)
with higher degree than m(t). Then, according to the division
algoritm, f(t) = q(t)m(t)
+ r(t), where r(t) has lower degree than m(t).
From this r(a) = 0, so r(t) is the zero
polynomial, and m(t) is a factor of f(t).
In an earlier example we proved that R[t]/<t^{2} + 1> is isomorphic to C. However, we do not need the complex numbers to prove that R[t]/<t^{2} + 1> is a field and to find the element j = t + <t^{2} + 1>. In fact, all properties of complex numbers can be translated into properties of elements of R[t]/<t^{2} + 1>. Since the construction of this field does not actually use C, we have here a method of defining the complex numbers algebraically. Since the subset {a + <t^{2} + 1>:a in R} is isomorphic to R, we have then defined C as an extension of R, and we have actually constructed this extension.
What we have just described is a particular case of a general procedure for construction of an extension of a field K. In the general case an irreducible polynomial over K will have a zero in the extension field. This is described in the following theorem by Kronecker.
If m(t) is a monic, irreducible polynomial over the field
K, then there is
an extension K(a):K, such that a has minimum polynomial m(t) over K. 
Before proving the theorem, we state the following. Let f and
g be coprime polynomials over a field. This means that they
have no common nonconstant factor. Then there exist polynomials a(t)
and b(t) over the same field, such that a(t)f(t)
+ b(t)g(t) = 1. This statement is analogous
to a theorem about integers, and we do not prove it.
Secondly, we show that K[t]/<m(t)> is an extension of K. Define a function F from K to K[t]/<m(t)> by F(x) = x + <m(t)> for all x in K. Then, if F(x) = F(y), x + <m(t)> = y + <m(t)>, i e x  y is in <m(t)>. Since x and y are elements of K this can only happen if x  y, as element of K[t]/<m(t)>, equals the zero polynomial. Then x = y, so F is injective. The requirements for a homomorphism are obviously fulfilled, so F is a monomorphism, and K[t]/<m(t)> is an extension of K.
Thirdly, let a = t + <m(t)>. Then a^{n} = t^{n} + <m(t)>, (k + <m(t)>)a^{n} = kt^{n} + <m(t)>, etc. From this, m(a) = m(t) + <m(t)> = <m(t)> = 0 in K[t]/<m(t)>, so a is a zero of m(t). For a polynomial n(t) of lower degree we get n(a) = n(t) + <m(t)> which is nonzero, since n(t) cannot be a multiple of m(t).
Finally, we show that K(a) = K[t]/<m(t)>.
K(a) consists of all rational expressions in t
+ <m(t)> with coefficients in K. All such
expressions can be written as polynomials in t + <m(t)>
of degree less than the degree of m(t). We do not show this
in detail, but the essential point is that f(t) + <m(t)>
has an inverse g(t) + <m(t)>
unless f(t) is a multiple of m(t). This is
shown as in part one of the proof.
Example 1: The polynomial t^{2} + 1 over R splits in R[t]/<t^{2} + 1>.
Example 2: Any polynomial over C splits in C. This statement is sometimes called the fundamental theorem of algebra. This name is also used for a couple of related statements, such as "any polynomial with complex coefficients has at least one complex zero".
Example 3: Over the field Z_{3} the polynomial t^{2} + 1 is irreducible since none of the three elements in Z_{3} is a zero. According to the divison algorithm any polynomial f(t) over Z_{3} can be written f(t) = q(t)(t^{2} + 1) + at + b, where a and b are elements of Z_{3}. Thus, any element in Z_{3}[t]/<t^{2} + 1> can be written at + b + <t^{2} + 1>. Since a and b take the values 0, 1, 2 independently of each other there are 9 combinations, so the field Z_{3}[t]/<t^{2} + 1> has exactly 9 elements. It is isomorphic to the so called direct product Z_{3}×Z_{3}.
V is a vector space over a field F if (V,+)
is an abelian group, and there is
defined a multiplication of elements in V with elements in F, such that
(ii) (a + b)u = au + bu for all a, b in F and u in V, (iii) a(bu) = (ab)u for all a, b in F and u in V, (iv) 1u = u for all u in V, (v) 0u = 0 for all u in V. 
If L:K is a field extension, then L is a vector space over K. The dimension of this vector space is called the degree of the extension, and is denoted [L:K].
If K(a):K is a simple algebraic extension, then
[K(a):K] equals
the degree of the minimum polynomial of a over K. 
This theorem shows why the degree n of the minimal polynomial is a kind of size measure for the extension field K(a), compared with K. A field K(a) with a larger n has greater size than a K(a) with a smaller n, in the same sense that an R^{n} with a larger n has greater size than an R^{n} with a smaller n.
Example 1: C = R(i) has a basis {1,i} as a vector space over R, so [C:R] = 2. Thus, the minimum polynomial of i over R has degree 2. This agrees, of course, with t^{2} + 1 being the minimum polynomial.
Example 2: [Q(2^{1/3}):Q] = 3 and [Q(sqrt(2)):Q] = 2, since the minimal polynomials are t^{3}  2 and t^{2}  2, respectively. Thus Q(2^{1/3}) is, so to speak, as much larger than Q(sqrt(2)), that R^{3} is larger than R^{2}, or a solid body is larger than a surface.
The great advantage of the connection between field extensions and vector
spaces is that it allows the theory of vector spaces, linear algebra, to
be used in the study of field extensions and their applications.
If L:K is a field extension, then a Kautomorphism
A of L is an automorphism of L, such that A(k)
= k for all k in K.
The Kautomorphisms of L form a group under composition
of functions, the so called Galois group of L:K, denoted
G(L:K). Under rather general conditions the number
of elements in G(L:K) equals [L:K].
However, the structure of the Galois group gives more detailed information
on the extension than the vector space structure.
A central theme in Galois theory is algebraic equations. Consider an extension K(a):K. While the dimension of K(a) as a vector space says, so to speak, how far a is from being in K, the Galois group says something on how a can be expressed by operations on the elements of K. Let S be a splitting field for f(t) over K. Then G(S:K) is isomorphic to a subgroup of the symmetric group S_{n}, where n is the degree of f(t). Further, the zeros of f(t) can be expressed by means of rational operations and root extractions on the coefficients of f(t) if and only if G(S:K) is a so called soluble group. Thus, the question of which algebraic equations can be solved by radicals is transformed into a problem of group theory.
The most famous example is certainly the fifth degree equation. The corresponding Galois group is a subgroup of S_{5}, sometimes the entire group. Now, it can be proved that S_{5} is not soluble, so fifth degree equations cannot, at least not always, be solved by radicals. However, S_{n} is soluble for all n < 5, so equations of degree up to 4 can be solved by radicals, as was of course known before the emergence of Galois theory.
If, in particular, K = Q we see that algebraic numbers are more complicated than the results of root extractions. Here we have a connection to algebraic number theory, too.
For those who want to go deeper into the subject, the book Galois Theory by Ian Stewart is recommended. Information on the more elementary parts of subject can be found in any textbook on abstract algebra and in mathematical encyclopedias.