# a split exact sequence of finite groups

Suppose G has a cyclic normal subgroup $\langle a\rangle$ of order $m$ and prime power index $s$ such that $m$ and $s$ are relatively prime. Then the following exact sequence splits:

$$1 \longrightarrow \langle a\rangle \longrightarrow G \longrightarrow G/\langle a\rangle \longrightarrow 1$$

Such group G is called a hyperelementary group.

Question: How to define a homomorphism $G/\langle a\rangle \rightarrow G$ to make the above sequence split ?

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Use Sylow's theorem. – Jack Schmidt Jul 7 '11 at 14:14
Yes. And the splitting works for all normal subgroups of $p$-power index which have order prime to $p$, whether they are cyclic or not. Rather deeper is the Schur-Zassenhaus theorem, which says that a normal subgroup whose order and index are relatively prime is complemented. – Geoff Robinson Jul 7 '11 at 14:37
oh, I see, Thanks! – Yubin Jul 7 '11 at 14:59