Let $|G| = 20$ and $G$ has only two elements of order $4$. Then $G$ is cyclic 
Let $|G| = 20$ and $G$ has only two elements of order $4$. Then $ G$ is cyclic.

I was trying to prove this assertion and I was given some hints also. But firstly I don't know about sylow theorems yet.

Definition: Let $G$ be a group and $|G|=p^n.q$, where $(p,q)=1$. Then every subgroup of order $p^n$ is called sylow $p$-subgroup.
Theorem (Sylow's III):
If $N_p$ is the number of sylow $p$- subgroup of a finite group $G$ ($|G|= p^n.q$, where $(p,q)=1$). Then $N_p = 1+kp$, where $k>=0$ and $N_p$ divide $q$. $( N_p\mid q)$.
Theorem: If there exists only one sylow $p$-subgroup for each prime $p$, which divides $|G|$. Then $G$ is direct product of those sylow $p$-subgroup.

My Attempt : Using these theorems I got to the point to prove that there's only one sylow $5$-subgroup and it is of order $5$.
Again, for sylow $2$-subgroups : $$ N_2 = (1+2k)\mid 5$$
Therefore,
$k = 0$ or, $2$
If $k=2$, then there's only $5$ subgroups of order 4 say $H_1,H_2,H_3,H_4,H_5$.
Now, given that there's only two elements of order 4.
Let, $|a|=4=|b|$
Then, $\langle a\rangle=\langle b\rangle $ otherwise resulting two more distinct elements $a^3,b^3$ of order 4.
Now here I came to a dead end. I know that if I can prove that $k=2$ results in contradiction then my proof is done. But I can't find any ways from here.
Any suggestions?
 A: Let $a$ be an element of order $4$. Then $\langle a \rangle$ is a subgroup of order $4$ so it is one of the Sylow $2$-subgroup.
Without loss of generality, assume that $H_1=\langle a \rangle$. Note that $H_1$ has two distinct elements of order $4$, namely $a$ and $a^3$.
Note that $H_1\cap H_2$ is a subgroup of $H_1$.
Case 1: $H_1\cap H_2=1$. Since $H_2$ is also a cyclic subgroup of order $4$, there exists an element of order $4$ in $H_2$ which is not $a$ or $a^3$; a contradiction.
Case 2: $H_1\cap H_2=H_1$. Then $H_1=H_1\cap H_2\leq H_2$. Since $|H_1|=|H_2|$, we have $H_1=H_2$; a contradiction.
Case 3: $H_1\cap H_2=\{1,a^2\}$. Since $H_2$ is cyclic group of order $4$, it has only one element of order $2$, so again there exists some element of order $4$ in $H_2$ which is not $a$ or $a^3$; a contradiction.
A: The existence of an element of order  $4$ means the Sylow-$2$ subgroups are cyclic.   Furthermore there's only one,  since otherwise we have more than two elements of order  $4$.
For the Sylow-$5$ subgroup, it's normal,  by the fact that the number of them is $1\pmod5$ and divides $4$.
We're done, since then $G $ is the direct  product $C_4\times C_5\cong C_{20} $ .
