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Let $G$ be a group and $H$, $K$ subgroups such that $H$ is normal in $G$ and $H$, $K$ are isomorphic. After some thought i intuitively concluded that $K$ is not necessarilly normal in $G$. Is that the case? Any rigorous argument? (e.g. counterexample)

Manos
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    My answer to this question contains counterexamples. As an exercise I invite you to find two subgroups of the dihedral group of 8 elements (= the symmetries of a square). Both are expected to be cyclic of order two. One central (hence normal) the other not. – Jyrki Lahtonen Sep 11 '11 at 20:05

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Yes. Take any group $G'$ with a non-normal subgroup $K'$ and consider $G = G' \oplus K'.$ Then $H = 1 \oplus K'$ is normal in $G$ and isomorphic to the non-normal subgroup $K$ obtained from embedding $K'$ in the first coordinate.

jspecter
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  • I like this direction, but i think it answers a different question. You are showing how to take a non-normal subgroup of a group and map it to an isomorphic subgroup of another group that is normal. In my question, i require that both subgroups, $H$, $K$ are subgroups of exactly the same group $G$. – Manos Sep 11 '11 at 20:26
  • Look again. Both $K$ and $H$ are subgroups of the group $G.$ – jspecter Sep 11 '11 at 20:28
  • @Manos: jspector has answered precisely your question. $H$ is the normal subgroup of $G$, and the other subgroup of $G$ is the one referred to at the end as being embedded in the first coordinate as $K'\oplus 1$. – Jonas Meyer Sep 11 '11 at 20:30
  • @jspecter: I now see. Beautiful :-) Thanks a lot! – Manos Sep 11 '11 at 20:36
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There are counterexamples with two elements in the group $\mathbb Z/2\mathbb Z\times S_3$.

Jonas Meyer
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