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I'm trying to prove that $G$ of order $35$ is Abelian by solely means of the Class Equation.

In general, if $|G|=pq$, with $p$ and $q$ distinct primes, then $|Z(G)|$ can in principle (Lagrange) be any among $1,p,q,pq$. If $|Z(G)|=p$, then every noncentral element has centralizer of order $p$, and hence the Class Equation reads:

\begin{alignat}{1} pq=p+kq \\ \end{alignat}

where $k$ is the number of noncentral conjugacy classes: contradiction, for $q\nmid p$. Likewise, if $|Z(G)|=q$, then we'd come up to:

$$pq=q+kp$$

which is again a contradiction, for $p\nmid q$. On the other hand, if $|Z(G)|=1$, then the Class Equation reads:

$$pq=1+k_pp+k_qq \tag 1$$

where $k_i$ are the number of conjugacy classes (i.e. orbits of the action by conjugation of $G$ on itself) of size $i$. In particular, for $p=5$ and $q=7$ the equation $(1)$ has only solution $(k_5,k_7)=(4,2)$; so there would be $4$ orbits of size $5$ and $2$ orbits of size $7$, and I expect that the orbit ("Class") equation $35=1+5+5+5+5+7+7$ lead to some contradiction. I know that the stabilizers (here centralizers) of the elements of one same orbit are all conjugate to each other, but I don't know whether this is relevant to come up to the contradiction I'm looking for.

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    Well, you have exactly $14$ elements of order $5$ (they are the ones in the orbits of size $7$). Since each subgroup of order $5$ contributes $4$ elements of order $5$, and two subgroups of order $5$ intersect trivially, how exactly do you manage to have exactly $14$ of them, if you can only add them four at a time? Similarly for order $7$, you have $20$ such elements, but you can only add them $6$ at a time, which is a bit of a problem. – Arturo Magidin Jun 28 '21 at 18:33
  • Thank you, @ArturoMagidin, I got the point. –  Jun 29 '21 at 18:25

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