Is $K$ Contained in the Center of $G$ if $|K|=5$ and $|G|$ is Odd?

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In summary, for a subgroup $K$ to be contained in the center of a group $G$, every element of $K$ must commute with every element of $G$. The order of a group, denoted by $|G|$, being odd is important in this question because it implies that the group has certain structural properties. However, it is possible for a subgroup of order 5 to not be contained in the center of a group of odd order. The size of the subgroup $K$ does not necessarily determine its containment in the center of $G$, and the fact that $|K|=5$ and $|G|$ is odd is not enough to determine whether $K$ is contained in the center of $G
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Euge
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MHB
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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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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 $K \subset Z(G)$.
 

FAQ: Is $K$ Contained in the Center of $G$ if $|K|=5$ and $|G|$ is Odd?

What does it mean for a subgroup to be contained in the center of a group?

When a subgroup is contained in the center of a group, it means that every element of the subgroup commutes with every element of the group. In other words, the subgroup is made up of elements that are fixed under all automorphisms of the group.

Why is the size of $G$ being odd significant in determining if $K$ is contained in the center?

The size of $G$ being odd is significant because it limits the possible orders of subgroups that can be contained in the center. Since $|G|$ is odd, the only possible orders for subgroups contained in the center are 1 and $|G|$. This allows for a simpler analysis of whether or not $K$ is contained in the center.

How does the size of $K$ being 5 affect its potential containment in the center of $G$?

The size of $K$ being 5 is significant because it is a prime number. This means that $K$ is a cyclic subgroup and its elements are all distinct powers of a single element. This simplifies the analysis of whether or not $K$ is contained in the center of $G$.

Is it possible for $K$ to be contained in the center of $G$ if $|K|=5$ and $|G|$ is even?

No, it is not possible for $K$ to be contained in the center of $G$ if $|G|$ is even. This is because if $|G|$ is even, the only possible orders for subgroups contained in the center are 1 and $\frac{|G|}{2}$, which do not include 5.

Can $K$ be contained in the center of $G$ if $|K|$ is not a prime number?

Yes, it is possible for $K$ to be contained in the center of $G$ if $|K|$ is not a prime number. However, this is a more complex analysis and depends on the specific orders of $K$ and $G$. In general, if $|K|$ is not a prime number, it is more difficult to determine if $K$ is contained in the center of $G$.

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