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Suppose $H$ is a subgroup of $G$. Is there always a surjective homomorphism $G\to H$?

I am confused with this. Today in my algebra class we had a group $G$ of order $p(p-1)$ for $p$ prime and the teacher said that because of Sylow's theorem $G$ contains a group of order $p$ (that part I totally agree) and hence there is a surjection $G\to\mathbb{Z}/p\mathbb{Z}$. Why is it so? Is it true in general or does this work for some reason specific to this example?

Edit: Actually the group was $(\mathbb{Z}/p^2\mathbb{Z})^\times$, but he stressed out that the only thing that matters is that it has order $p(p-1)$. So this is why I asked this as a general question for groups of order $p(p-1)$. Thanks to the answers, its not true (which releases a little my confusions). But then is true in this case?

Spenser
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2 Answers2

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For a counterexample, let $p=3$, and let $G$ be the group of permutations of a $3$-element set (the unique non-abelian group of order $6$). This $G$ has no normal subgroup of order $2$, and therefore has no surjection to a $3$-element group.

Andreas Blass
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    It might be worth noting that if we have a subgroup $C\simeq C_p$ of order $p$ in $|G|=p(p-1)$ and a surjection $\eta:G\to C_p$, then $K=\ker \eta\lhd G$ has order $p-1$ coprime to $C$, so $C\cap K=1$ and $G$ is a semidirect product $K\rtimes C$. In this case, since $S_3$ has a normal $3$-Sylow subgroup $A_3 \simeq C_3$, the semidirect product would be direct, but $S_3$ is not abelian. This is the smallest counterexample, I think. – Pedro Mar 20 '14 at 01:18
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Also, you can let $G=A_5$ be the alternating group on $5$ letters. This group is simple, so its only normal subgroups are $0$ and $A_5$. The first isomorphism theorem implies that the image of any group morphism from $A_5$ has order $60$ or $1$. Therefore, $A_5$ cannot surject onto any subgroup except for itself and the trivial subgroup.

Peter Crooks
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