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If $H \trianglelefteq G$, need $G$ contain a subgroup isomorphic to $G/H$?

I worked out the isomorphism types of the quotient groups of $S_3, D_8, Q_8$.

For $S_3$:

  1. $S_3/\{1\} \cong S_3$,
  2. $S_3/\langle (1\ 2\ 3)\rangle \cong \mathbb Z_2$,
  3. $S_3/S_3 \cong \{1\}$.

For $D_8$:

  1. $D_8/\{1\} \cong D_8$,
  2. $D_8/\langle r\rangle \cong \mathbb Z_2$,
  3. $D_8/\langle s, r^2\rangle \cong \mathbb Z_2$,
  4. $D_8/\langle sr^3, r^2\rangle \cong \mathbb Z_2$,
  5. $D_8/\langle r^2\rangle \cong V_4$,
  6. $D_8/D_8 \cong \{1\}$.

For $Q_8$

  1. $Q_8/\{1\} \cong Q_8$,
  2. $Q_8/\{1, -1\} \cong V_4$,
  3. $Q_8/\langle i \rangle \cong \mathbb Z_2$,
  4. $Q_8/\langle j \rangle \cong \mathbb Z_2$,
  5. $Q_8/\langle k \rangle \cong \mathbb Z_2$,
  6. $Q_8/Q_8 \cong \{1\}$.

So I'm guessing that the statement is true, but I don't know how to prove it. And if its not true, I haven't found a counter example. Can someone give me a proof or counterexample? Or a HINT :D

EDIT: Ahhh. I feel stupid now. Given $\{1, -1\}$ normal in $Q_8$ there is no subgroup of $Q_8$ isomorphic to $V_4$. Correct? So the statement is false?

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  • $\begingroup$ correct.$\phantom{.}$ $\endgroup$ – Julian Kuelshammer Oct 21 '12 at 6:48
  • $\begingroup$ That G has a normal subgroup H says that G is an extension of H. We say that it is a split extension if there is a map G/H->G compatible with the "short exact sequence" H->G->G/H (in the sense we have a commutative diagram), in which case G has a subgroup isomorphic to G/H and we say G is a semidirect product of H and G/H. In general, extensions need not split. $\endgroup$ – anon Mar 10 '13 at 19:39
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No. This statement is false. Observe that $\{1, -1\} \trianglelefteq Q_8$ and $Q_8/\{1, -1\} \cong V_4$ but no subgroup of $Q_8$ is isomorphic to $V_4$.

(Immediately realized this right after posting, sorry!).

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The statement not true in general, as Robert showed. But it does hold for finite abelian groups (since all subgroups of finite abelian groups are normal, we can drop the normal condition). You should be able to convince yourself of this by showing it for $\mathbb{Z}_n$ and then noting that any finite abelian group is a direct sum of cyclic groups.

To see that the word "finite" is necessary, just consider $2\mathbb{Z}\leq\mathbb{Z}$.

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