I am studying for my algebra qualifying exam in January, and I have a question regarding the proof I have for this question in Hungerford, problem 13 under the Sylow Theorems section.
The way I approached it is as follows:
By the third Sylow theorem, the number of Sylow p-subgroups divides $|G|$ and is of the form $pk+1$. The divisors of $p^2$ are $1$, $p$ and $p^2$. However, the only divisor of the form $pk+1$ is $1$ (when $k=0$). So, $G$ only has one Sylow p-subgroup, and therefore, by corollary 5.8, this Sylow p-subgroup has order $p^2$ and is normal. Meaning, the Sylow p-subgroup is the entire group $G$, and since it is normal, it is therefore abelian.
Is this correct, or do I need to use the hint like suggested? Thank you!
Are you suggesting that a normal subgroup is always abelian? I don't see how you are getting abelian from normal.
Here's a hint for a more typical solution to this problem: consider possible orders of $Z(G)$, and recall that for $p$-groups one has $Z(G)\neq\{1\}$.