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Let [tex]\phi: G \rightarrow G'[/tex] be an isomorphism of a group [tex]<G, *>[/tex] with a group [tex]<G', *'>[/tex]. Write out a proof to convince a skeptic of the intuitive clear statement.

Problem:

41.) If H is a subgroup of G, then [tex]\phi[H] = {\phi(h)| h \in H}[/tex] is a subgroup of [tex]G'[/tex]. That is, an isomorphism carries subgroups into subgroups.

Thoughts on the problem:

I am not sure what is mean by "an isomorphism carries subgroups into subgroups." Nevertheless, I went to get help to draw up a proof for this problem. After coming up with a proof, I reviewed it a little later, and was a little skeptical about the homomorphism aspect (shown below).

Proof:

Let [tex]x, y \in \phi(H).[/tex]

Now, Suppose [tex]\exists a, b \in H:

x = \phi(a),

y = \phi(b).[/tex]

Therefore,

[tex]xy = \phi(a) \phi(b) = \phi(ab) \in \phi(H).[/tex] (#1)

Also, [tex]e = \phi(e) \in \phi(H),[/tex] (since the identity is in G, the subgroup H must have the same identity). (#2)

Finally, [tex]x^{-1} = [\phi(a)]^{-1} = \phi(a^{-1})[/tex] shows there is an inverse element in the subgroup H. (#3)

Questions:

In equation (#1), can we conclude that [tex]\phi(a) \phi(b) = \phi(ab)?[/tex] It seems kind of vague to me but all I've gathered from the problem is that our function [tex]\phi[/tex] takes the group G with its binary operator * to another group G' with an operator *'. Neither operators (*, *') are defined, so how can we conclude [tex]\phi(a) \phi(b) = \phi(ab)?[/tex]

In equation (#2), and equation (#3) the reasoning seems fairly straightfoward. In these two equations we've shown that the identity [tex]e \in H[/tex] is also in the group G'. Similarly we've shown that the same inverse in H (which is a subgroup of G), is also found in the group G'.

So what I'm really puzzled about is how the first equation (equation (#1)), contains/proves the homomorphism property (to obtain the closure condition). We've shown that [tex]xy = \phi(ab) \in \phi(H)[/tex] fulfills closure, this is useful since we were trying to prove the subgroup condition. However, in doing so we had to show the homomorphism property to get there. To me the whole proof is pretty good except I just can accept that [tex]\phi(a) \phi(b) = \phi(ab)[/tex].

Thanks,

JL

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# Isomorphism (Fraleigh 7th: Section 5)

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