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Problem of the Week #204 - April 26, 2016

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Euge

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Jun 20, 2014
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Euge

MHB Global Moderator
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Jun 20, 2014
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This week's problem was answered correctly by Deveno . You can read his solution below.



Let $G$ be a group. If $\text{Aut}(G)$ is cyclic, then so is any subgroup of $\text{Aut}(G)$.

In particular, $\text{Inn}(G)$, the subgroup of inner automorphisms, is cyclic.

But $\text{Inn}(G) \cong G/Z(G)$, where $Z(G)$ is the center of $G$.

If $G/Z(G)$ is cyclic, $G$ is abelian (the following is the standard proof):

Let $xZ(G)$ be the generator of $G/Z(G)$. Then any $g \in G$ is of the form $x^kz$, with $k \in \Bbb Z$, and $z \in Z(G)$.

So given $g,h \in G$, we have:

$gh = (x^kz_1)(x^mz_2) = x^kx^m(z_1z_2) = x^{k+m}z_1z_2 = x^{m+k}z_2z_1 = x^mx^k(z_2z_1) = (x^mz_2)(x^kz_1) = hg$.
 
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