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Problem of the Week #358 - May 4, 2020

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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
Here is this week's POTW:

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Suppose $K$ is a normal subgroup of a group $G$ with $|G|$ odd. Prove that if $|K| = 5$, then $K$ is contained in the center of $G$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,892
No one answered this POTW. You can read my solution below.

Let $G$ act on $K$ by conjugation. Since $G$ is odd, every stabilizer has odd index. Thus the class equation for $K$ is either $5 = 1 + 1 + 3$ or $5 = 1 + 1 + 1 + 1 + 1$. Since $K$ has prime order, $K \cap Z(G)$ is either trivial or all of $K$. This eliminates the former possibility, so that the class equation is $5 = 1 + 1 + 1 + 1 + 1$; in particular, $K\cap Z(G) = K$, or $K \subset Z(G)$.
 
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