I have a question about the following problem [Finite Group Theory, Martin Isaacs, Chapter 5]:
Let $ B $ and $ C $ be cyclic subgroups of a finite group G, and suppose that $ BC = G $ and $ B \cap C > 1 $. Assume that $ C \lhd G $ and that $ |C:G^{'}|=n $. Show that the order of $ M(G) $ is strictly less than $ n $.
I report my reasoning:
For the third theorem of homomorphism, I know that $ B/B \cap C \cong BC/C=G/C $ and since $ B $ is cyclical, every quozient is also cyclical. So I'm in the hypothesis of the exercise 5A.5 from M.Isaacs Finite Group Theory and then I have to $ |M(G)| $ divides $ |C:G^{'}|=n $. Therefore $ |M(G)| $ is less than $ n $.
I can't justify the fact that it has to be strictly less than $ n $. (I think it depends on the fact that $ C $ is a normal subgroup proper to $ G $).
Let $\Gamma$ be a Schur representation group for $G$. So $\Gamma$ has a subgroup $Z \cong M(G)$ with $Z \le Z(\Gamma) \cap \Gamma'$ and $\Gamma/Z \cong G$.
Let $\bar{C}$ and $\bar{B}$ be the inverse images of the subgroups $C$ and $B$ of $G$ in $\Gamma$. So $\bar{B} \cap \bar{C}$ is the inverse image of $B \cap C$. Choose $\bar{b}\in\bar{B}$ and $\bar{c}\in\bar{C}$ as inverse images of generators of $C$ and $B$, so $\Gamma = \langle \bar{b},\bar{c},Z\rangle$. Then $\bar{B} \cap \bar{C}$ is centralized by $\bar{b}$, $\bar{c}$ and $Z$, so $\bar{B} \cap \bar{C} \le Z(\Gamma)$.
Note also that $\Gamma'$ is the inverse image of $G'$, so $|\bar{C}:\Gamma'| = |C:G'| = n$. Also $\bar{C}/Z \cong C$ is cyclic, so $\bar{C}$ is abelian.
Now we can apply Lemma 4.6 of Isaacs' book to the group $\Gamma$ with the abelian normal subgroup $\bar{C}$ of $\Gamma$. This tells us that $|\bar{C} \cap Z(\Gamma)| = |\bar{C}:\Gamma'| = n$. Since $\bar{B} \cap \bar{C} \le Z(\Gamma)$, and $Z$ is a proper subgroup of $\bar{B} \cap \bar{C}$, $|Z| =|M(G)|$ divides $n$ and is less than $n$.