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Let $G$ be a group with of order 170 and $|Z(G)|$ divisible by 2. Prove that $G$ is cyclic.

I'm thinking that I need to use Sylow's theorem in some way but I don't really know where to start.

Appreciate any help!

giantsandmath13
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1 Answers1

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The order of $G/Z(G)$ is a divisor of $85$, so it is $1$, $5$, $17$ or $85$. The first case means $Z(G)=G$.

The cases $5$ and $17$ cannot happen, because if $Z(G)$ is proper the quotient $G/Z(G)$ cannot be cyclic.

See Groups of order $pq$ are cyclic for the case $85$: such a group is cyclic, because $5$ does not divide $17-1$. By the same argument as before, this case can be dismissed.

Thus $G$ must be abelian. Can you finish?

egreg
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  • @giantsandmath13 If $gZ(G)$ is a generator, then every element of $G$ is of the form $g^kx$ for $x\in Z(G)$. Now $(g^kx)(g^ly)=(g^ly)(g^kx)$ because $x,y\in Z(G)$, so the group is abelian, which is impossible due to $Z(G)\subsetneq G$. – egreg Apr 27 '17 at 21:56