Dtyjlui

Here is an incomplete proof:

If $Hle G$ is a subgroup, there are only three possibilities for its cardinal: $1,p$ or $p^2$ by Lagrange.

If $|H|=p^2$ then there exists $gin Hsubset G$ of order $p^2$ and so $G$ is cyclic.

If $|H|=1$ and there is no other proper subgroup then $G$ is cyclic.

So there must be a subgroup of order $p$, let $|H|=p$. Let $x$ be in $G$. We will show that $xHx^{-1}=H$ and so $Htriangleleft G.$ By contradiction let's suppose $xHx^{-1}ne H$.

$xHx^{-1}$ is a subgroup of $G$ because $$(xhx^{-1})^{-1}=xhx^{-1}in xHx^{-1} text{ and }~ xhx^{-1}xh'x^{-1}=xhh'x^{-1}in xHx^{-1}~forall h,h'in xHx^{-1}$$

There is a formula for the cardinality of a product of groups that says $$begin{align}|xHx^{-1}|cdot|H|&= |xHx^{-1}H|cdot|xHx^{-1}cap H|\ |(xHx^{-1})H| &= frac{|xHx^{-1}|cdot|H|}{|xHx^{-1}cap H|}=frac{pcdot p}{1}end{align}$$

And here I'm not sure how to show that $xHx^{-1}$ has cardinality $p$, I guess for the same reason as $H$.

And why is the intersection trivial?

If the equality above is true we get that $xHx^{-1}H=G$ so for any $xnotin H$ there exist some $h_1,h_2in H$ such that $x=xh_1x^{-1}h_2$

$$begin{align}e &=h_1x^{-1}h_2\ h_1^{-1}&= x^{-1}h_2\ h_1^{-1}h_2^{-1}&= x^{-1}implies xin H text{ (impossible)}end{align}$$

Once we arrive at a contradiction we have that $xHx^{-1}=H$ so $H$ is normal in $G$.

Then I have to show that there exists $Kle G$ such that $Kcap H={e}$ and $Ktriangleleft G$ and from this it will be easy to show that $Gcong Ktimes H$

You can argue like this.

If there were an element of $G$ if order $p^2$, then $G$ were cyclic, generated by this element; thus, every non-identity element of $G$ has order $p$. Fix arbitrarily a non-identity element $g_1in G$, and then an element $g_2in G$ not lying in the subgroup generated by $g_1$. For $iin{1,2}$, denote by $G_i$ the subgroup generated by $g_i$. Both these subgroups are cyclic; hence, generated by any of their non-identity elements. It follows that the intersection of $G_1$ and $G_2$ is trivial (any non-identity element of the intersection would generate both $G_1$ and $G_2$, but $G_1ne G_2$). It follows that all the $p^2$ products $h_1h_2$ with $h_1in G_1$ and $h_2in G_2$ are pairwise distinct. Thus, $G$ is exactly the set of all these products, as wanted.

$H$ has cardinality $p$. The map $Hto xHx^{-1}$ defined by $hto xhx^{-1}$ is one to one and onto. I'll leave you to check that, it is very easy. Hence $xHx^{-1}$ must have the same cardinality. As for the intersection there are only two options: it is either of size 1 or $p$. If it has size $p$ then it means that $H$ and $xHx^{-1}$ is the same set which contradicts your assumption.

What is $H$? Define your variables! I assume that what you mean is something along the lines of: take any non-identity element $h$ of $G$, and let $H$ be the subgroup that it generates.

This subgroup is cyclic, so cannot be all of $G$, as $G$ is not cyclic, and is not trivial, as it contains $h$, so must have order $p$.

Now take any $g in Gsetminus H$. Then $langle g, hrangle$ contains $H$, but is strictly larger than it, and is a subgroup of $G$, so must be all of $G$. Thus, $G$ is generated by $g$ and $h$. Now, the order of $g$ must also be $p$ (it must divide $p^2$, can't be $1$ since $g$ is not the identity, and can't be $p^2$ since $G$ is not cyclic). Let $K$ be the subgroup generated by $g$. Since $H$ and $K$ both have order $p$ and don't coincide, their intersection must be trivial. Thus, we merely need to show that both are normal in $G$. But given that we chose $H$ to be an arbitrary order-$p$ subgroup, it suffices to show it for $H$.

This follows from your argument: $y mapsto xyx^{-1}$ is injective (if $xyx^{-1} = xzx^{-1}$, then multiplying that equation by $x^{-1}$ on the left and $x$ on the right gives $y = z$), and is surjective when restricted to map from $H$ to $xHx^{-1}$ by definition of the latter, so $|xHx^{-1}| = |H| = p$. The intersection is trivial since it's a subgroup of $H$, so has order $1$ or $p$, but isn't all of $H$, so can't have order $p$.

Thus, $H$ and $K$ are normal, and $G = H times K$.