So far, my attempt of proof looks like this:
$\#G =45=3^2\times 5$, so:
- By the 1st Sylow Theorem that G has p-subgroups of order 5, 9 and 3.
By the 3rd theorem, the p-subgroups of order 5 and 9 have to be unique: $$n_5 = 1 \text{ (mod } 5), n_5 |9 \implies n_5=1$$ The proof is analogous for $P_9$ and so they're both normal in G (by adding the 2nd Theorem, all the p-subgroups are conjugate and there can't be any conjugate of $P_5$ or $P_9$ that isn't the group itself, since we've already shown they're unique).
For every p-subgroup, we know it's center is not trivial, so $\#Z(G)>1$
Now, I'm having trouble linking the information about its p-subgroups to the equation: $$\#G = \# Z(G) + \sum_{i=1}^\infty[G:N(x_i)]$$ I feel like there's something quite obvious I'm missing but I can't put my finger on it.
I've looked into this post also, but the example they give has references to a "Corolary 15" to show that the center has either order 6 or 48, but the theorem isn't there so idk what theorem they're refering too.