# $H$ normal in $G$. Need $G$ contain a subgroup isomorphic to $G/H$

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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correct.$\phantom{.}$ – Julian Kuelshammer Oct 21 '12 at 6:48
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. – anon Mar 10 at 19:39

## 1 Answer

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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You're right --- – Martin Brandenburg Oct 21 '12 at 7:32
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