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Suppose that G is a finite group, and that H and K are normal subgroups of G with trivial intersection, and suppose that H and K are isomorphic. Is it true that the quotient groups G/H and G/K are isomorphic?

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2 Answers

Let $G=\mathbb Z_2\times\mathbb Z_4$. Find two subgroups in $G$ isomorphic to $\mathbb Z_2$ and intersecting trivially such that the quotients of $G$ by them are not isomorphic.

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Thank you. I realize that H and K must somehow be conjugate to each other in order for the respective quotient groups to be isomorphic. Is this enough? I think there should be an automorphism from G to itself that takes H to K. –  Iota May 23 '11 at 2:09
    
@Iota: indeed. (This is exactly what I said in my answer...) –  Pete L. Clark May 23 '11 at 2:14
    
@Iota: or to put Pete's answer into high-faluting sounding words, it's enough that $H$ and $K$ be conjugate in the holomorph $G\rtimes \mathrm{Aut}(G)$ of $G$. –  Arturo Magidin May 23 '11 at 2:32
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As Mariano has shown, the answer is a clear no.

The best repair I can think of is the following: suppose that $H$ and $K$ are normal subgroups of a group $G$ such that there exists an automorphism $\varphi: G \rightarrow G$ with $\varphi(H) = K$. Then $G/H \cong G/K$: indeed, the isomorphism is induced by $\varphi$. In particular, the above condition holds if $H$ and $K$ are conjugate subgroups of $G$, which is useful enough to be worth remembering.

Note finally that the condition that $H \cap K = \{e\}$ seems to have nothing to do with anything: it neither helps nor hurts the desired conclusion, so far as I can see.

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Thank you. What if there are no automorphisms of G that sends H to K, does that suffice to conclude that the quotient groups cannot be isomorphic? Hmmm...There is a natural projection from G to G/H, and also from G to G/K, in order to have a well-defined map from G/H to G/K, we must have H to be in the kernel of the second projection to G/K. But since H and K have trivial intersection, H cannot be in the kernel, so there is no well-defined map between the quotient groups so no isomorphism. Does this make sense? –  Iota May 23 '11 at 2:39
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@Iota: No, that is not sufficient. The maximal subgroups of the dihedral group of order 8 need not even be isomorphic, but the quotient groups are all cyclic of order 2. In larger examples (order 32 and 48), one can have isomorphic H,K that are both characteristic (so not conjugate in Aut(G) to any other subgroup), where G/H and G/K are isomorphic. –  Jack Schmidt May 23 '11 at 4:42
    
I don't mean to be redundant, but I just wanted to add, for people who didn't see it right away (like me), that to prove $G/H \cong G/K$, given the conditions set by Pete above, you consider the surjective homomorphism $\pi_K \circ \varphi \colon G \to G/K$ whose kernel is $H$. Thanks everyone for your great postings! –  Dan Douglas Aug 25 '13 at 7:03
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