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Let $G$ be a group of order $20$ in which the conjugacy classes have sizes $1$, $4$, $5$, $5$, $5$. Then state whether true or false

A) "$G$ contains a normal subgroup of order $4$".

The answer is supposed to be false. But I don't think so. Since one of the conjugacy classes is of size $5$ I assume some element has a centralizer of order $4$. Since $$|cl(a)|= \frac{|G|}{|C(a)|} \ ,|cl(a)|=size \ of \ conjugacy \ class \ of \ a,\ |G|=order\ of \ group, \\|C(a)|=\ order \ of \ centralizer \ of \ a.$$

And since the centralizer is a normal subgroup isn't the group supposed to have a normal subgroup of order $4$.

B) "$G$ contains a subgroup of order $10$."

This is supposed to be true. Can anyone give a reason why?

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We have a group $G$ whose class equation is $1+4+5+5+5=20$.

(A) Every normal subgroup is a disjoint union of conjugacy classes. Each normal subgroup contains the conjugacy class of order $1$, since it contains the identity. A non-trivial normal subgroup must contain at least one other conjugacy class. The other conjugacy classes have sizes $4$, $5$, $5$, $5$. Hence, the order of a non-trivial normal subgroup must be at least $5$.

(B) Note that a group of order $20$ has a unique Sylow-$5$ subgroup $N$. Also note that a group of order $20$ must have an element $x$ of order $2$. Let $H$ be the subgroup generated by $x$. Note that $HN$ is a subgroup of $G$, since $N$ is normal in $G$, and $HN$ is order $10$, since $H$ and $N$ intersect trivially.

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    $\begingroup$ @Bungo Thanks for catching that! $\endgroup$ – user729424 Jan 23 '20 at 2:38
  • $\begingroup$ How do you know that N and H have only one common element ? $\endgroup$ – Siddharth Prakash Feb 1 '20 at 9:07
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    $\begingroup$ The order of an element always divides the order of the group. Since $N$ has order $2$ and $H$ has order $5$, any element in common would have an order that divided both $2$ and $5$. So any common element would have to have order $1$. So the only common element must be the identity. $\endgroup$ – user729424 Feb 1 '20 at 17:00
  • $\begingroup$ Thank you for your help $\endgroup$ – Siddharth Prakash Feb 1 '20 at 20:18
  • $\begingroup$ No problem! Thanks for the fun problem! $\endgroup$ – user729424 Feb 1 '20 at 20:18
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The centralizer is normal in the normalizer, but needn't be normal in $G$.

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  • $\begingroup$ Another true statement is " G contains a normal subgroup of order 5". Can you please explain why it is true $\endgroup$ – Siddharth Prakash Jan 23 '20 at 1:31
  • $\begingroup$ The Sylow $5$-subgroup is normal, mind you, because there's only one. $\endgroup$ – Chris Custer Jan 23 '20 at 2:07
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    $\begingroup$ More on the first part: from the class equation, $G$ is not abelian. If the Sylow $2$-subgroup were normal, $G$ would be the product of its Sylow subgroups. Any group of order $4$ is abelian. So $G$ would be abelian. Contradiction. $\endgroup$ – Chris Custer Jan 23 '20 at 2:14

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