Suppose that $H$ is a subgroup, of order $p$ (a prime), of a group $G$, of order $< p^2$ . Show that $H\trianglelefteq G$.
I know that $|G| = np$ where $n \in \mathbb{N} \text{ and } n < p$. I thought I should make a homomorphism $\phi:G \to gH$ such that $\phi(a) = a(gH)$.
The kernel of this function $\text{Ker}\phi = \{g \in G |\forall a \in G, gaH = aH \} = \{ g \in G| \forall a^{-1}gaH = H \}$
\begin{align} \text{Ker}\phi &= \{g \in G |\forall a \in G, gaH = aH \} \\ &= \{ g \in G|\forall a \in G,a^{-1} gaH = H \}\\ &= \{ g \in G| a^{-1}ga \in H\} \\ &= \bigcap_{a \in G} aHa^{-1} \le H \end{align}
However I am not sure how to procced from here.
Here is another solution I have just come up with:
We know that $$ |PP^g| = |PP^g| = \frac{|P||P^g|}{|P \cap P^g|} = \frac{p^2}{|P \cap P^g|} $$ Since the above fraction has to divide evenly, this must mean that $|P \cap P^g| = p, |P \cap P^g| = 1$ or $|P \cap P^g| = p^2$.
- The second case cannot occur as if $|P \cap P^g| = 1$ then $|PP^g|=p^2$ which is bigger than the order of $G$, which is at most $p^2 -1$.
- We cant have $|P \cap P^g| = p^2$ as the largest $|P \cap P^g|$ can equal is $p$.
- If $|P \cap P^g| = p$ then we have that $P = P^g$ since both $P$ and $P^g$ are both subgroups with order $p$ and the order of the intersection of both these groups is also $p$.
Since $P = P^g$ then $P\trianglelefteq G$.
for an automorphism $f:G\to G$ and $h\in H$ it has to be $f(h)\in H$ since $H$ is the only subgroup of $G$ with order $p$ and $o(f(h))|o(h)=p$