# Isomorphic quotient groups

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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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
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.
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