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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.

hsalin
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3 Answers3

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

Swapnil Tripathi
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  • 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
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    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
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    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
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    Haha. "you don't see the forest because of all the trees.." You made me smile. :) – Swapnil Tripathi Oct 20 '14 at 19:50
  • @Math83 - no problem, you learned something I guess! – Nicky Hekster Oct 20 '14 at 20:08
  • @NickyHekster: Sorry, I thought it was the OP. – Swapnil Tripathi Oct 20 '14 at 20:10
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$|G|=p^k$ 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 $|C(x)|$

Dhruv Joshi
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Bhaskar Vashishth
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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.

Ari Royce Hidayat
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