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I am confused about the following question, it looks obvious but turns out hard for me.

If group G can be embedded into another group H, prove that there exists a group S that is isomorphic to H, such that G≤S.

(G can be embedded into H means there exists an isomorphism between G and a subgroup K of H.)

I tried to use the union of G and H\K to make S, and expand the isomorphism between G and K to the one between H and S, but I met the difficulty defining the operation on that union when considering the possible existence of common elements of G and H\K.

I find the operation I defined does not satisfy the law of union properly and if there are common elements, the union may not be enough.

But if I make S without using H, it may be tough to define the isomorphism between S and H.

Perhaps my thinking is completely wrong.

How can S and the isomorphism between H and S be constructed? Or any other ways to solve it without construction?

Thank you for any assistance.

user1426842
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  • If you are worried about the possibility that the sets $G$ and $H \mid K$ are not disjoint, then step 1 would be to first find a group $H'$ and an isomorphism $H \mapsto H'$ such that $G$ and $H'$ are disjoint. This is a purely set theoretic issue. – Lee Mosher Oct 08 '24 at 18:26

2 Answers2

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The disjoint union $S=G\sqcup (H\smallsetminus K)$ is a good idea. I assume you realize there is a bijection $f:S\to H$. Use this bijection to define the group structure! In other words, define

$$s*s'=f^{-1}(f(s)*f(s')).$$

Because you have defined $f$ using the embedding $G\hookrightarrow H$ and inclusion $H\smallsetminus K\hookrightarrow H$, this successfully combines the group structures coming from $G$ and $H\smallsetminus K$ in the way you're hoping for.

What's left is to convince yourself this is actually a valid group law which makes $S\simeq H$, and to see that $G$ is a subgroup of $S$. A minor technical point here: make sure that the group law on $G$ is actually the restriction of the group law on $S$!

TY Mathers
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  • Perhaps taking the disjoint union of $G$ and $H \setminus K$ would be better, given the OP's expressed concern about the possibility that they are not disjoint to begin with. – Lee Mosher Oct 08 '24 at 21:38
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You're blurring the lines between isomorphic and different groups. I avoid doing this except in certain specific cases, like different subgroups of the same group, for instance, the four different Klein four subgroups of $S_4,$ or different Sylow subgroups of a given group. Or I guess if I were to dabble in representation theory.

Another way:

it suffices to find a non-trivial automorphism of $H.$

Any non-trivial group has a non-trivial automorphism. See here.

suckling pig
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