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### Proof of Subgroup Involving an Isomorphism

```Date: 11/04/2004 at 14:25:11
From: Bridget
Subject: Isomorphisms acting on Groups

Suppose ö is an isomorphism from a group G to a group G'.  Prove that
if K is a subgroup of G, then ö(K) = {ö(k) | k is in K} is a subgroup
of G'.

We know that K < G.  I know to show that a group is a subgroup you
must use the two step process of showing that it is closed under the
operation ö and closed under the inverse, but I'm not sure how to
start this.  Could you help me out?

```

```
Date: 11/05/2004 at 18:09:26
From: Doctor Jordan
Subject: Re: Isomorphisms acting on Groups

Hi Bridget,

Let us express H = o(K).  Recall that for H to be a subgroup of G', it
must be a nonempty subset of G' such that H is closed under the
operation of G', each h in H has an inverse contained in H, and the
identity element of G' is contained in H.  We are given that K is a
subgroup of G, hence it contains the identity element of G, each
element in it has an inverse, and it is closed under the operation of G.

Since o is an isomorphism, it is a homomorphism, and for e, y in K a
homomorphism must satisfy o(ey) = o(e)o(y).  In particular, for e is
the identity element of K and thus ey = e, then o(e) must also be the
identity element of G', as otherwise o(e)o(y) would not equal o(y).
Thus we know that H has the identity of G'.

By assumption, all elements of H are of the form o(x) for x in G.  Let
us have two arbitrary elements of H, x and y.  By definition of o,
they are of the form o(x') = x, o(y') = y for x', y' in K.  But since
K is a subgroup it is closed and so x'y' in K, hence o(x'y') =
o(x')o(y'),  which shows that the product of any two elements of H is
in H.

Can you show that for any element x in H, its inverse is also in H?
It is helpful to remember that since K is a subgroup, each element in
K has a (unique) inverse in K.

If you have any problems with this or any other questions, please
write me back!

- Doctor Jordan, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Modern Algebra

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