Show that a group of order $28$ contains $2$ subgroups $H_1 > H_2$ such that $|H_1|= 14$ and $|H_2| = 7$ I know that there is only one $7$-Sylow group in $G$ ($n_7= 1$). 
I just need to find a normal subgroup of order $2$ , I know that there is an element of order $2$ in $G$ but cannot continue.
 A: Hints: Let $H_2$ be that Sylow $7$ subgroup.


*

*What can you say about the quotient group $G/H_2$?

*Apply the 1-1 correspondence between subgroups of $G/H_2$ and the subgroups of $G$ containing $H_2$.

A: Let $H_2$ be $7$-Sylow subgroup of $G$. As you know, $H_2$ is a normal subgroup. Consider $\frac{G}{H_2}$. Since every group of order $p^2$ is an abelian group where $p$ is prime, $\frac{G}{H_2}$ is abelian and so $G'\subseteq H_2$.
Now, assume that $H_1$ is a maximal subgroup which contains $H_2$. Since $G'$ is contained in $H_1$, $H_1$ is a normal subgroup. Hence, $[G:H_1]=p$ where $p$ is a prime number. Note that $[G:H_1][H_1:H_2]=[G:H_2]$. We can conclude that $|H_2|=14$.
A: I thought I would add an answer that shows that you were essentially already done.
You have just one $7$-Sylow subgroup, call it $H_2$ (as it is the subgroup of order $7$ you need). So $H_2$ is normal in $G$.
You also have an element of order $2$, call it $x$. Let $H = \{e,x\}$ where $e$ is the identity element of $G$. Now $H_1 = \langle x,H_2\rangle = HH_2$ is a subgroup of order $14$ since $H_2$ is normal.
A: $H_2$ is normal, then consider $G':=G/H_2$, we have $|G'|=4=2^2$. Then by Sylow's first theorem $G'$ contains an element $g$ of order $2$. Let $K':=\langle g\rangle$, we have $|K'|=2$. Moreover, $K'$ is of the form $H_1/H_2$ where $H_1$ is a subgroup of $G$ which contains $H_2$ and $|H_1|=14$
