# 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.

## 2 Answers

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 $1,p,p^2,...p^n$

• Hmm, since $C(x)$ is a subgroup, its order must divide $|G| = p^n$, thus it must be a power of $p^k$ where $k<n$. But how can you conclude that for any $x$ the number of cosets of its centralizer $[G:C(x)]$ also is a prime power? If $x \not\in Z(x)$ and $[G:C(x)] = 1$ then this is false since by definition $ax = xa$ holds for $C(x)$ for some $a \implies a \in C(x)$ as well, so $[G:C(x)]$ must be at least 2? – hsalin Oct 20 '14 at 19:44
• The [index=number of cosets] of a subgroup is a divisor of the order of the whole group! – Nicky Hekster Oct 20 '14 at 19:45
• $|G:C(x)|=|G|/|C(x)|=\frac{p^n}{p^k}$ where $k<n$ as you say. – Swapnil Tripathi Oct 20 '14 at 19:48
• Oh my god, @NickyHekster , this is what I totally missed, it's like "you don't see the forest because of all the trees..". Of course $[G:C(x)]$ divides the group order, that is why the order must be some value of $1,p,p^2,...,p^n$.. Thanks! – hsalin Oct 20 '14 at 19:49
• @SwapnilTripathi : yes, I just realized that now, thanks a lot! – hsalin Oct 20 '14 at 19:49

$|G|=p^k$ for some $k$ as it is a $p$ group,()we are only talking finite here, this statement may not hold in 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 $|C(x)|$

• Very underrated answer. – IAmNoOne Apr 16 '18 at 7:37