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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  is an isomorphism from a group G to a group G'.  Prove that
if K is a subgroup of G, then (K) = {(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  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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