Every group of order $p^2$, $p$ prime is isomorphic to $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_{p}\oplus \mathbb{Z}_{p}$
I am confused about two parts of this proof.
Proof:Assume every nonidentity element of this group $G$ has order $p$. Then $\langle a\rangle$ is normal otherwise there is an element $b$ in $G$ such that $bab^{-1} \notin \langle a\rangle$
Here is my first source of confusion. if $\langle a\rangle$ is not normal I would suspect there is an element $b$ in $G$ with $ba^{i}b^{-1} \notin \langle a\rangle,\text{for some}\space i \in \mathbb{Z}$. Why does $bab^{-1} \notin \langle a\rangle$ necessarily hold with $a$?
Next part of confusion
Since $\langle a\rangle \cap \langle bab^{-1}\rangle =\{1\}$ the distinct left cosets of $\langle bab^{-1}\rangle$ are $\langle bab^{-1}\rangle,a\langle bab^{-1}\rangle,…,a^{p-1}\langle bab^{-1}\rangle$
is this because there must be $p$ distinct cosets and there union must be $G$, so this must be all of the cosets ?
Best Answer
Alternate proof: The center of a $p$-group is nontrivial. Then $G/Z(G)$ is cyclic $\implies G$ is abelian.