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Let $G_1$ and $G_2$ be two abelian groups such that $G_1 \approx G_2$ ($G_1$ is isomorphic to $G_2$). $H_1$ is a subgroup of $G_1$ and $H_2$ is a subgroup of $G_2$ such that $H_1 \approx H_2$. Check whether ${G_1}/{H_1} \approx {G_2}/{H_2}$. What if $G_1$ and $G_2$ were finite groups?

The last part (finite case) is where I need help.

I first tried to define a function (a potential isomorphism) $\phi$ from ${G_1}/{H_1}$ to ${G_2}/{H_2}$ using the isomorphism, say, $\phi_G$ from $G_1$ to $G_2$ as $\phi(gH_1)=\phi_G(g)H_2, \;g \in G_1$.

For $\phi$ to be well defined we need that $\phi_G(a^{-1}b) \in H_2$ whenever $a^{-1}b \in H_1$, where $a,b \in G_1$, i.e., $\phi_G(H_1) \subseteq \phi_G(H_2)$, which need not always be true.

I know that this was just a hit and trial relation which isn't even always a function. At this point I was told the answer that it's not true, that is, the factor groups need not be isomorphic and there are examples which I was then asked to find.

I found that ${\mathbb{Z}}/{2\mathbb{Z}} \not \approx {\mathbb{Z}}/{4\mathbb{Z}}$ while rest of the conditions hold.

Now comes the last part. I wonder if we could extend the isomorphism from $H_1$ onto $H_2$ to $G_1$ onto $G_2$ by mapping the elements of $G_1\setminus H_1$ to $G_2\setminus H_2$ in a certain way in the finite case and use this isomorphism to create one between the factor groups. I would like to see some counterexample if it is false for finite groups too.

Aman Kushwaha
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  • Any help is appreciated. This question was asked in an interview. – Aman Kushwaha Jul 06 '23 at 07:07
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    While everything about quotient groups, from the words to the notation we use, is meant to invoke associations to division, it's not very common to use actual fractions to denote them. It is much more common to see $G/H$ than to see $\frac GH$. And writing $G\setminus H$ is certainly wrong, as that can only mean the (set-theoretic) complement of $H$ in $G$ (all the elements of $G$ which are not in $H$). – Arthur Jul 06 '23 at 07:10
  • @Arthur I changed the factor group notation. By $G\setminus H$ I mean the set theoretic complement only. – Aman Kushwaha Jul 06 '23 at 07:14
  • Oh, you actually meant to talk about the complement. That's fine, then. It's a common mistake, so I assumed it was like that here too. I'm sorry. If I had actually read that paragraph a bit more carefully, I think I would've known. – Arthur Jul 06 '23 at 07:16
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    @Arthur $(\mathbb Z_2\times\mathbb Z_4) / \langle (0,2) \rangle \approx K_4$ whereas $(\mathbb Z_2\times\mathbb Z_4) / \langle (1,0) \rangle \approx Z_4$. Is that correct? Here "$\times$" means external direct product. – Aman Kushwaha Jul 06 '23 at 07:44
  • Yes, it is. It is the example you would stumble upon if you tried to find the smallest example you could. "I want the quotients to be non-isomorphic, so they must have order at least 4. The full group needs to be a multiple of that, so at least 8. There are three abelian groups of order 8, and the cyclic one only has a single subgroup of order 2 so that can't work, and $\Bbb Z_2^3$ doesn't have order 4 elements so that can't work. Let's try the last one." – Arthur Jul 06 '23 at 07:48
  • Yeah, this is the smallest indeed. Thank you so much @Arthur – Aman Kushwaha Jul 06 '23 at 07:49

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