How to show that $|G|=d_1^2+\cdots +d_s^2$ and $G$ is abelian?

Question

In Representation Theory of Finite Groups: An Introductory Approach by Benjamin Steinberg, there is a corollary in which I feel confused.

Corollary 6.2.6. Let p be a prime and let $G$ be a group of order $p^2$. Then $G$ is abelian.

Proof. Let $d_1, \cdots,d_s$ be the degrees of the irreducible representations of $G$. Then $d_i$ can be $1$, $p$ or $p^2$. Since the trivial representation has degree $1$ and $$p^2=|G|=d_1^2+\cdots+d_s^2$$ it follows that all $d_i=1$ and hence $G$ is abelian.

I have two questions:

1. Why does the equation $p^2=|G|=d_1^2+\cdots+d_s^2$ hold?
2. Why is $G$ abelian if each of its irreducible representations has the degree $d_i=1$?

I have not yet found any theory before Corollary 6.2.6 in this book to illustrate these questions. Could you please give me some hints to show them?

Answer 1

For any finite group, the sum of the squares of the dimensions of its irreducible representations is the order of the group - i.e. the equation $$\vert G\vert = d_1^2 + \ldots + d_s^2$$ is true for any finite group, and since $\vert G\vert = p^2$ we get the full equation. This follows from character theory and the orthogonality relations.

For your second question, I'd have a think about the conjugacy classes, and how the number of conjugacy classes of a group is related to the number of irreducible representations - how many irreducible representations does $G$ have if each $d_i=1$?

Let me know if you want any more hints or explanations :)