Group with at least 2 subgroups of order $p$ has at least $p^2-1$ elements of order $p$.

Question

Let $G$ be a finite group and $p$ be a prime number. Let $a,b$ be two elements of order $p$ such that $b\notin \langle a\rangle $ where $\langle a\rangle $ denotes the subgroup generated by $a$.

Prove that $G$ has at least $p^2-1$ elements of order $p$.

Now $\langle a\rangle $ will have $p-1$ elements of order $p$ and since $b\notin \langle a\rangle $ if we consider $\langle b\rangle $ then we also have $p-1$ elements of order $p$. Thus $G$ has $2(p-1)$ elements of order $p$.

But I will have to prove that there are $p^2-1$ elements of order $p$.

Please give some hints to complete the proof

Answer 1

Hint: think about what will be the order of $ab$, where $a\in \langle a \rangle$ and $b\in \langle b \rangle$

Answer 2

If $a$ and $b$ commute they generate a subgroup isomorphic to $Z/p \oplus Z/p$ which contains $p^2-1$ elements of order $p$.

The other option is that repeated conjugation by $a$, applied to subgroup $\langle b \rangle$, creates $p$ isomorphic and distinct copies of that subgroup. The copies have $p^2-p$ elements of order $p$. The additional $(p-1)$ elements of order $p$ come from $\langle a \rangle$ which cannot have nontrivial intersection with the copies.

We have broken the symmetry between $a$ and $b$ here. Can more [than $p^2-1$ ] elements of order $p$ be produced by using $b$ to conjugate the group generated by $a$? It looks true in the non-commuting case.