A group of order $n^2$ with $n+1$ subgroups of order $n$ with trivial intersection is abelian The following problem is from my exercise sheet in group theory. I managed to do the first two items, but i am stuck with the third. Hints and comments on the solutions presented below will be the most apreciated.
Problem:
Let G be a group of order $n^2$ with $n+1$ subgroups of order $n$, such that the intersection of any two of them is $\{ e \}$. Show that:
a) If $H$ and $K$ are two subgroups of order $n$, then $HK = G$.
b) If $H$ is a subgroup of order $n$, then $H$ is normal in G.
c) G is abelian.
Solution:
a) Since $H$ and $K$ are finite subgroups of order $n$ and $H \cap K = \{e\}$, we have that
$$
|HK| = \frac{|H||K|}{|H \cap K} = \frac{n.n}{1}= n^2
$$
Then, since $HK \subset G$ and have the same number of elements, then $HK = G$.
b) Let $K \leq G$ such that $|K| = n, \ K \not= H$. By the previous item, we know that  $KH=G$. Since the left cosets form a partition of the set, we have that 
$$
G = \bigcup_{k \in K}kH.
$$
Now, suppose that for some fixed $k \in K, kH \not= Hk$. Since both sets have the same number of elements, then 
$$
Hk \setminus kH \not= \emptyset \  \therefore \  \exists \ k' \in K \setminus
\{k\},\  \exists \ h,h' \in H; \ hk = k'h'
$$ 
Then
$$
k' = hk(h')^{-1} \in (eH)(kH) = kH
$$
what is a contradiction. Therefore, $kH = Hk \ \forall \ k \in K$ and then follows that $H \vartriangleleft G$.
c) ?
Thanks in advance.
 A: Let $H,K$ be two distinct subgroup of order $n$. Now $\mid HK\mid=\frac{\mid H\mid \mid K\mid}{\mid H\cap K\mid}=\mid H\mid \mid K\mid=n^2=\mid G\mid \Rightarrow G=HK$
For normality of $H$, we have to show that , $gHg^{-1}=H,\forall g\in G$. At the contrary let there is some $g\in G$ such that $gHg^{-1}\ne H$. Since, $\mid gHg^{-1}\mid =n$ , from the first part we get that $gHg^{-1}H=G$.So there is some, $h_1,h_2\in H $ such that $g=gh_1g^{-1}h_2\Rightarrow g\in H\Rightarrow H=G$ , which is impossible. So $H\trianglelefteq G$
Now let $H_1,H_2,....,H_{n+1}$ be the list of all n-ordered subgroups of $G$. Then $\mid H_1\cup H_2\cup ....\cup H_{n+1}\mid =(n-1)(n+1)+1=n^2\Rightarrow G=H_1\cup H_2\cup ....\cup H_{n+1}$.Now let $g_1\in G$. So, there is some n-ordered subgroup $H_i$ such that $g_1\in H_i.$ Consider any $g_2\in H_j, H_i\ne H_j$. Now $g_1g_2g^{-1}_1g^{-1}_2\in H_i\cap H_j$, using the normality of $H_i,H_j\Rightarrow g_1g_2g^{-1}_1g^{-1}_2={e_G}\Rightarrow g_1g_2=g_2g_1$. So $\mid C(g_1)\mid \ge n^2-(n-1)$ . But as $n\ge 2$ $C(g_1)$ forces to be $G$. As $g_1$ was chosen arbitrarily, we can say $ G$ is commutative.
A: That $G$ is abelian can be shown by the following counting argument. Apparently, $G = \bigcup_{i=1}^{n+1}H_i$, where $H_i$ are the subgroups of order $n$. Let $x \in G$, say $x \in H_i$ for some $i$. Then, the centralizer of $x$, $C_G(x)$ contains at least $n(n-1)+1$ elements, since all the elements outside $H_i$ commute with $x$ (Why is this? Use the fact that, in general, if two normal subgroups $H$ and $K$ have trivial intersection, then $[H,K] \subseteq H \cap K=1$, whence all the elements of $H$ commute with those of $K$ and vice versa). Hence for the conjugacy class $$|G:C_G(x)|=|Cl_G(x)| \leq \frac{n^2}{n^2-n+1}.$$But one can easily show that $\frac{n^2}{n^2-n+1} \lt 2$ for all $n$. Hence $|Cl_G(x)|=1$ for all $x \in G$, which is equivalent to $G$ being abelian.
