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 $S_3$:
- $S_3/\{1\} \cong S_3$,
- $S_3/\langle (1\ 2\ 3)\rangle \cong \mathbb Z_2$,
- $S_3/S_3 \cong \{1\}$.
For $D_8$:
- $D_8/\{1\} \cong D_8$,
- $D_8/\langle r\rangle \cong \mathbb Z_2$,
- $D_8/\langle s, r^2\rangle \cong \mathbb Z_2$,
- $D_8/\langle sr^3, r^2\rangle \cong \mathbb Z_2$,
- $D_8/\langle r^2\rangle \cong V_4$,
- $D_8/D_8 \cong \{1\}$.
For $Q_8$
- $Q_8/\{1\} \cong Q_8$,
- $Q_8/\{1, -1\} \cong V_4$,
- $Q_8/\langle i \rangle \cong \mathbb Z_2$,
- $Q_8/\langle j \rangle \cong \mathbb Z_2$,
- $Q_8/\langle k \rangle \cong \mathbb Z_2$,
- $Q_8/Q_8 \cong \{1\}$.
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 😀
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?
Best 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!).