# Is it necessary that $f: G_1 \to G_2$ is an isomorphism, for $f:H\to H$ an automorphism with $G_1, G_2\le H$ of the same cardinality?

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.

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.

• It's worth noting that this is consistent with my answer. – Shaun Apr 22 at 21:02
• 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. – Ufomammut Apr 22 at 21:03

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.

• Oh no, I forgot to say in my original question that the two subgroups have same cardinality. – Ufomammut Apr 22 at 20:44
• I've edited the answer, @Ufomammut. – Shaun Apr 22 at 20:47
• What is your automorphism here? – Mark Apr 22 at 20:47
• 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. – Shaun Apr 22 at 20:52
• 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 $f:G_1\to G_2$. – Mark Apr 22 at 20:52