Let $G_1,G_2$ be groups with 2 subgroups respectively $H_1,H_2$ satisfying certain conditions, must $|G_1:H_1|=|G_2:H_2|$

Let $G_1,G_2$ be groups with two subgroups respectively $H_1,H_2$ such that there is a bijection $f:G_1\rightarrow G_2$ and $f|H_1$ is a bijection between $H_1,H_2$. Must $|G_1:H_1|=|G_2:H_2|$ ?

Note: If we only require that $G_1,G_2$ have the same size and $H_1,H_2$ have the same size, then it does not follow that $|G_1:H_1|=|G_2:H_2|$. As a counterexample, set $H_1=G_1=G_2=\mathbb{Z}\times\mathbb{Z},H_2=\mathbb{Z}\times\{0\}$

Thank you

• Is $f$ a homomorphism, or merely a bijection of sets? If $f$ is a bijection of sets, it looks like you've already answered your own question, more or less. If $f$ is a homomorphism, then $f$ is an isomorphism and all structure (including the relevant indexes) will carry over. Commented Nov 8, 2013 at 21:07
• @BrettFrankel The examples indicate OP is talking about infinite groups. In principle the equality might fail. (As evidence, take for example the fact that it's possible for an automorphism of an infinite group $G$ to deflate a subgroup $H< G$ into an even smaller subgroup, a proper subgroup of $H$!)
– anon
Commented Nov 8, 2013 at 21:08
• @anon Yes, I'm aware of examples, but don't see how they constitute any obstruction. If $f$ is an isomorphism, it will induce a bijection on the sets of cosets, so there isn't much to say here unless I'm really missing something. And if $f$ is just a bijection, then the OP has already exhibited a counter-example. Commented Nov 8, 2013 at 21:17
• @BrettFrankel Are you sure the map between coset spaces is injective? It seems like it'd be possible for the preimage of $H_2$ under $f$ to strictly contain $H_1$, in which case there are nontrivial cosets of $H_1$ mapping to the trivial coset $H_2$. | Edit: oops, no, if $f$ is injective then $H_1$ is exactly the preimage of $H_2$ and so this is trivial. Oh well.
– anon
Commented Nov 8, 2013 at 21:18
• @anon The OP has specified that $f$ restricts to a bijection between $H_1$ and $H_2$. Commented Nov 8, 2013 at 21:21

1 Answer

Take a bijection between

$H_1=\mathbb{Z}\times {0}$ and $2 \mathbb{Z} \times 2\mathbb{Z} =H_2$ then extend this bijection to a bijection of

$\mathbb{Z}\times \mathbb{Z}\rightarrow \mathbb{Z} \times \mathbb{Z}$

you can do that since in the complement there are infinite countable elements