suppose we have a group G and a subgroup P, s.t. |P|=p ( a prime ) and $|G|<p^2$. I want to show that P is normal in G.
My attempt :
So i did this question for the more specific case of $|P|=7, |G|=28$ and was able to prove it for that case so i will try the same method as I did for that.
P is normal in G $\Leftarrow \Rightarrow, \forall p \in P, \forall g \in G ; gpg^{-1} \in P$
now consider $PP^g$
$P^g=\{g^{-1}pg|p\in P\}$, if we define a function $f(p)=g^{-1}pg$ it is easy to show by right and left cancellation that if $g^{-1}xg=g^{-1}yg$ then x=y and so the function is one to one which implies that $|P|=|P^g|$
So $P^g$ also has order p.
$P <G, P^g<G$
so we can say that
$|PP^g|=\tfrac{|P||P^g|}{|P\bigcap P^g|}=\tfrac{p^2}{|P\bigcap P^g|}$
Now what is $P \bigcap P^g$.
First note that $\forall H_1,H_2 that are subgroups of G$, then $H_1 \bigcap H_2$ is a subgroup of G. therefore $P\bigcap P^g$ is a subgroup of G.
It also follows that $H_1\bigcap H_2$ is a subgroup of both $H_1$ and $H_2$ so $P\bigcap P^g$ is a subgroup of P.
$\Rightarrow |P\bigcap P^g| \|P|$ $\Rightarrow |P\bigcap P^g| \ p$
So $|P\bigcap P^g|=1$, or $7$
But as $P \neq P^g$ and $|P|=|P^g|$ it follows that P is not contained in %P^G% so $P \bigcap P^g \neq P$ so $P \bigcap P^g $ is contained in P.
$\Rightarrow |P \bigcap P^g|<|P|$
$\Rightarrow |P \bigcap P^g|=1$
$\therefore |PP^g|=p^2$
Therefore $PP^g$ is not a subgroup of G and so P is the only subgroup of order 7. this means that $P=\{g^-1pg|p \in P\}$ which proves that P is normal in G.
Is this correct ?
Your last sentence is confusing. A better explanation would be this: you assumed $P\ne P^g$ and got that $|PP^g|=p^2$ which is not possible because $PP^g$ is a subset of $G$ which has less than $p^2$ elements. A contradiction. Hence $P$=$P^g$ for all $g\in G$ and that means $P\triangleleft G$.
Everything else in your proof is good, though if you are familiar with the Sylow theorems you could make it much shorter. $|G|=pm$ when $m<p$. Then a subgroup of order $p$ is a $p$-Sylow subgroup of $G$. By Sylow theorems $n_p|m$ and $n_p\equiv 1$(mod $p$). But as $m<p$ you have no choice, it must be $n_p=1$. Hence there is only subgroup of order $p$, so it must be normal in $G$.
I'm sorry if you don't know the Sylow theorems yet, just if you do then this proof is much easier.