# Centralizers and conjugacy classes

Let $$G$$ be a finite group with order 168 such that for each element $$g \in G$$ wih order $$7$$, the centralizer $$C_G (g)$$ is equal to $$\langle g \rangle$$. Compute the number of conjugacy classes of elements $$g$$ of order $$7$$ and find the size of the conjugacy class for such an element.

I tried to begin with the class equation, but didn't have any luck. I don't know if I will need to use the definition of a Sylow subgroup at some point here.

I would appreciate any help.

Let $$G$$ act on $$G$$ by conjugation. Then each orbit of an element is its conjugacy class, and the size of the orbit is the index of the element's stabilizer. We're given that the stabilizer (which is the element's centralizer) has size $$7$$ so it has index $$24$$. Thus, each conjugacy class of an element of order $$7$$ has size $$24$$.
The $$7$$-Sylow subgroups of $$G$$ are not normal. If one were, its non-trivial elements would have a conjugacy class of size no more than $$6$$.
Thus, $$G$$ must have $$8$$ $$7$$-Sylow subgroups (because no other divisor $$d$$ of $$168$$ satisfies $$d \equiv 1 \pmod 7$$). Two distinct $$7$$-Sylow subgroups of $$G$$ must have trivial intersection, which means $$G$$ has $$48$$ distinct elements of order $$7$$. Therefore, there are $$2$$ such conjugacy classes.
• Thank you for the very helpful answer. The only remaining thing I have a question on is this: by the Sylow theorems, I understand that there are either $n_7 = 1$ or $n_7 = 8$ $7$-Sylow subgroups. If $n_7 = 1$, then $G$ is normal, but I cannot understand how to rule this out. Could you explain more about this? – user860403 Dec 11 '20 at 5:54
• @StanMS Check my revised answer. If $N= \langle g \rangle$ were a normal $7$-Sylow subgroup of $G$, then every conjugate of $g$ would necessarily lie within $N$ so the size of its conjugacy class would be no larger than $6$, and we know that's not true. Therefore, $N$ can't be normal. – Robert Shore Dec 11 '20 at 7:53
• I think I understand: a $7$-Sylow subgroup has an identity (order 1) and six other elements. If it's normal, it's closed under conjugation, so all its conjugates have to live in that group, but we have only six more slots. The only other question I have is: how do we know there's an element of order $7$? Is it by Lagrange's theorem, which states that the order of an element is either $1$ or $7$, and we already have the identity? – user860403 Dec 11 '20 at 8:06
• Yes, any Sylow subgroup has to have elements of prime power order, and there can only be a single element of order $1$. – Robert Shore Dec 11 '20 at 8:25
• Thank you. But last question: you wrote $N = \langle g \rangle$. Why must $N$ be generated by a single element $g$? Is this central to the proof? – user860403 Dec 11 '20 at 8:32