Part of simple proof of nontrivial center in p-group I'm trying to understand the proof of a Burnside theorem (as stated in Beachy's Abstract Algebra p. 328): Let $p$ be prime number. The center of any $p$-group is nontrivial.
Now, In the proof they say that if we let $G$ be a $p$-group, then in the class equation $$|G| = |Z(G)|+\sum [G:C(x)]$$
for all $x$ that is not in the center and represent a conjugacy class, we see that every term in $\sum [G:C(x)]$ is divisible by $p$ since $x\not\in Z(G) \implies [G:C(x)]>1$. This last statement is what I do not understand, how do we know that $p \mid [G:C(x)]$ for any conjugacy class?
I know that the elements in the conjugacy class of $x$ is in bijection with the cosets of $C(x)$, i.e. $[G:C(x)]$, but how can we be certain that the number of elements in a conjugacy class of $x$/cosets of the centralizer of $x$ is divisible by $p$?
Best regards.
 A: Hint: If $x$ is not in the center, then what contradiction would you get if $|G:C(x)|=1$.
Note: the values $|G:C(x)|$ can take are $1,p,p^2,...p^n$
A: $|G|=p^k$ for some $k$ as it is a $p$ group, we are only talking about finite groups here, this statement may not hold for infinite groups.
Now as $C_G(x)<G$ therefore $|C(x)|$ divides $p^k \implies |C(x)|=p^i$ for some $0 \le i < k$, so, $|G:C(x)|=p^{k-i}$ and $k-i>0$ implies $p$ divides $|C(x)|$
A: Let $P$ be the p-group, by class equation:
$|P| = |Z(P)| + \displaystyle\sum_{i=1}^r |P:C_P(g_i)|$
where $g_1, g_2, ..., g_r$ be representatives of distinct conjugacy classes of $P$ not contained in the center $Z(P)$.
Write the class equation above as:
$|P| = p^\alpha = a + b$ where $a = |Z(P)|$ and $b = \displaystyle\sum_{i=1}^r |P:C_P(g_i)|$.
By definition $C_P(g_i) \ne P$ for $i = 1, 2, ..., r$ so $|P:C_P(g_i)| = b \ne 1$. Because $b$ is the index of $C_P(g_i)$ in $P$, $b$ must be in the power of $p$, so $p$ divides $b$. Because $p$ obviously divides $p^\alpha$, and $p$ divides $b$, $p$ must also divide the remaining term $a$, so $a = |Z(P)|$ cannot equal $1$. Hence $Z(P)$ is not trivial.
