# (Normal) subgroups of different orders of a group of order $20.$

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?

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.

• @Bungo Thanks for catching that! – user729424 Jan 23 '20 at 2:38
• How do you know that N and H have only one common element ? – Siddharth Prakash Feb 1 '20 at 9:07
• 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. – user729424 Feb 1 '20 at 17:00
• Thank you for your help – Siddharth Prakash Feb 1 '20 at 20:18
• No problem! Thanks for the fun problem! – user729424 Feb 1 '20 at 20:18

The centralizer is normal in the normalizer, but needn't be normal in $$G$$.

• Another true statement is " G contains a normal subgroup of order 5". Can you please explain why it is true – Siddharth Prakash Jan 23 '20 at 1:31
• The Sylow $5$-subgroup is normal, mind you, because there's only one. – Chris Custer Jan 23 '20 at 2:07
• 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. – Chris Custer Jan 23 '20 at 2:14