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This might be a very trivial question, and I have explained what I think about it below. Let's say I have an automorphism $f : H \to H$. Now, let's say I take two subgroups $G_1$ and $G_2$ of $H$. Is it necessary that $f: G_1 \to G_2$ is also an isomorphism?

I think yes, cause all the elements of $G_1$ and $G_2$ are still the elements of H, and if $f$ is an automorphism, $f: G_1 \to G_2$ should necessarily be an isomorphism too.

Edit: Originally, I forgot to say that the two subgroups $G_1$ and $G_2$ have same cardinality.

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  • $\begingroup$ Please try to make your titles more informative. That, other users can benefit from your questions more readily. $\endgroup$ – Shaun Apr 22 at 20:55
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If you're assuming that $\text{Im}(f|_{G_1}) \subseteq G_2$ (otherwise the question doesn't really make any sense), then the answer is yes if $G_1$ is finite and not necessarily otherwise.

If $G_1$ is finite, then since $f$ is monomorphism, $f : G_1 \to G_2$ is also a monomorphism and a monomorphism between two finite groups of the same cardinality is an isomorphism.

If $G_1$ can be infinite, we can take $H = \bigoplus_{n \in \mathbb{N}} \mathbb{Z}, \ G_1 = \{0\} \ \oplus \ \bigoplus_{n \geq 1}\mathbb{Z}, \ G_2=H$ and $f = 1_H$. Then $f : G_1 \to G_2$ is not surjective but $G_1$ and $G_2$ have the same cardinality.

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  • $\begingroup$ It's worth noting that this is consistent with my answer. $\endgroup$ – Shaun Apr 22 at 21:02
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    $\begingroup$ I just found a note online similar to what I wanted to understand. In their example, the map is $f(h) = mhm^{-1}$ for some $m \in H$ and their $G_2$ was $mG_1m^{-1}$. So, I think the assumption that $\text{Im}(f|_{G_1}) \subseteq G_2$ is satisfied in that case. $\endgroup$ – Ufomammut Apr 22 at 21:03
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No.

Let $G_1=\Bbb Z_6$, $G_2=D_{6}$, and $H$ be some suitably large $S_n$, $f=1_H$.


NB: Here $D_6$ is the dihedral group of six elements.

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  • $\begingroup$ Oh no, I forgot to say in my original question that the two subgroups have same cardinality. $\endgroup$ – Ufomammut Apr 22 at 20:44
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    $\begingroup$ I've edited the answer, @Ufomammut. $\endgroup$ – Shaun Apr 22 at 20:47
  • $\begingroup$ What is your automorphism here? $\endgroup$ – Mark Apr 22 at 20:47
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    $\begingroup$ Because $D_6$ (with $6$ elements) is isomorphic to $S_3$ and, for sufficiently large $n$, there is a cyclic subgroup generated by an element of order six and $S_3$ is isomorphic to a subgroup, @Ufomammut. $\endgroup$ – Shaun Apr 22 at 20:52
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    $\begingroup$ They are isomorphic to subgroups of $S_n$ for some large $n$, this follows from Cayley's theorem. Though here elements of $\mathbb{Z_6}$ are just not mapped to elements of $D_6$. I think there is a problem in the statement of the question. If you look at $f:G_1\to f(G_1)$ then it is an isomorphism. If you take any two random subgroups then not sure we can even talk about $f:G_1\to G_2$. $\endgroup$ – Mark Apr 22 at 20:52

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