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?
Best Answer
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.