Class Equation : $20 = 1 + 4 + 5 + 5 + 5$, normal subgroup of order $5$ not normal subgroup of order $4$ I have the following class equation: 
$$20 = 1 + 4 + 5 + 5 + 5$$
I know that there are subgroups of order $4$ and $5$ in $G$. I see this because $\vert G \vert = \vert \text{centralizer} \vert \cdot \vert \text{conjugacy class} \vert$ and the centralizer is a subgroup of $G$. So, we have $\vert \text{centralizer} \vert = 4$ and $\vert \text{centralizer} \vert = 5$, so there are subgroups of order $4$ and $5$ in $G$. However, I have been told that the subgroup of order $4$ is not normal and the subgroup of order $5$ is normal in $G$. I know that a normal subgroup is the union of conjugacy classes. Is there a way I can easily show their normality? Thank you.
 A: Use Sylow-Theorem to show normality.$|G|=20=2^2*5$.Hence to get $n_5,1+5k/4,for k=0,1,2...$this means $n_5=1$.Hence G has normal subgroup of order 5.
A: The centralizers of the elements of the only conjugacy class of size $4$ have order:
$$|C_G(g)|=\frac{|G|}{\left|\operatorname{cl}(g)\right|}=\frac{20}{4}=5$$
In general $\langle g\rangle\le C_G(g)$, so in this case ($5$ prime and Lagrange) $\langle g\rangle= C_G(g)$. Therefore, there are precisely $4$ elements of order $5$, which then must be grouped into one single subgroup of order $5$, thence normal.
On the other hand, you'll not be able to build up a subgroup of order $4$ as the union of the unique central singleton (the set with the identity element) and some other noncentral conjugacy classes (in fact, all of them have size $4$ or $5$). So, a group of order $4$ (which exists because the centralizers of the elements of the conjugacy classes of size $5$ have order $4$) is not normal.
A: First of all it's fairly trivial that any subgroup that is a union of conjugacy classes is normal (and conversely).
You can read off from the class equation that the union of the center,  which is evidently trivial,  and the conjugacy class of order $4$ can give a normal subgroup of order $5$.  (That doesn't mean it does;   but it might. )
Again,  the center is the conjugacy class of size $1$.  That's clearly the identity.   It needs to be in any subgroup.   So as you can see there's no way to put a $1$ together with a $4$ or a $5$ and get a $4$.  Thus no normal subgroup of order $4$.
So by looking at the class equation you can see what size normal subgroups are possible. That's you can rule some out.   But for instance in this case we need to check whether the subgroup of order $5$ is indeed normal.
The only possible normal subgroups are of order $5$ and $10$.  The other possibilities are ruled out by Lagrange and the class equation.
A: If there are two subgroups of order $5$ say $H$ and $K$ then $o(HK)=o(H)o(K)=25>20$. Therefore there can only be one subgroup of order $5$ which implies it is a normal subgroup.
Let $G$ contain a normal subgroup $M$ of order $4$. Then $G$ is isomorphic to $M×H$. This implies $G$ is abelian. This contradicts the given class equation.
Therefore subgroup of order $4$ cannot be normal.
