Page 78 - DMTH403_ABSTRACT_ALGEBRA

Unit 6: Group Isomorphism

Note that a homomorphism f from G to G carries the product x * y in G to the product Notes
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f(x) * f(y) in G .
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Note The word ‘homomorphism’ is derived from two Greek words ‘homos’,
meaning ‘link’, and ‘morphe’, meaning ‘form’.

Let us define two sets related to a given homomorphism.
Definition: Let (G , * ) and (G , * ) be two groups and f : G  G be a homomorphism. Then we
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define
(i) the image of f to be the set
Im f = {f(x) 1 x  G }.
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(ii) the kernel of f to be the set
Ker f = {x  G | f(x) = e }, where e is the identity of G .
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Note that Im f  G , and Ker f = f ({e }  G .
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Example: Consider the two groups (R, +) and (R*,.). Show that the map exp : (R, +)  (R*,.)
: exp(r) = e is a group homomorphism. Also find Im exp and Ker exp.
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Solution: For any r , r  R, we know that e 1 r  2 r  e .e .
2 r
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:. exp(r + r ) = exp(r ).exp(r ).
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Hence, exp is a homomorphism from the additive group of real numbers to the multiplicative
group of non-zero real numbers.
Now, Im exp = {exp(r) | r  R} = {e | r  R},
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Also, Ker exp = {r  R | e = l} = {0}.
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Note that examples takes the identity 0 of R to the identity 1 of R*. example also carries the
additive inverse – r of r. to the multiplicative inverse of exp (r).
Example: Consider the groups (R, +) and (C, +) and define f : (C, +)  (R, +) by f(x + iy) =
x, the real part of x + iy. Show that f is a homomorphism. What are Im f and Ker f?
Solution: Take any two elements a + ib and c + id in C. Then,
f((a + ib) + (c + id)) = f((a + c) + i(b + d)) = a + c = f(a + ib) + f(c + id)

Therefore, f is a group homomorphism.
Imf = {f(x + iy) | x, y  R } = { x | x  R ) = R.
So, f is a surjective function

Ker f = { x + iy  C | f ( x + iy ) = 0 } = { x + iy  C | x = 0 }
= { iy | y E R }, the set of purely imaginary numbers.
Note that f carries the additive identity of C to the additive identity of R and ( – z) to – f(z), for any
z  C.

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