Calculation of the Schur Multiplier (Problem 5A.8(b) Isaacs' Finite Group Theory)

Question

Let $A$ and $B$ be arbitrary finite groups.

a. Show that $$|M(A \times B)| \geq |M(A)||M(B)|.$$

b. Assuming the $|A|$ and $|B|$ are coprime, show that $$M(A \times B) \cong M(A) \times M(B)$$

(where $M(G)$ denotes the Schur multiplier of the finite group $G$).

I answered part $a$ by constructing a central stem extension of $A \times B$, from the direct product of the Schur representation groups for $A$ and $B$.

I am stuck on part b. I'll detail my approach below.

Let $\Gamma$ be a Schur representation group for $A \times B$. Then there is a surjective map $\pi : \Gamma \to A \times B$ such that $\Gamma / ker\;\pi \cong A \times B$, and $ker\; \pi \subseteq Z(\Gamma) \cap \Gamma'$. Let $A^* = \pi^{-1}(A)$ and $B^* = \pi^{-1}(B)$. So then $G/A^* \cong B$ and $G/B^* \cong A$, since $|\Gamma:A^*|$ and $|\Gamma:B^*|$ are coprime it follows that $\Gamma = A^* B^*$. Then $|\Gamma| = |A^*||B^*|/|A^* \cap B^*|$. Since $\Gamma / A^* \cong B$ it follows that $|B^*|/|A^* \cap B^*| = |B|$, hence $|A^* \cap B^*| = |ker \; \pi|$. Since $ker \; \pi \subseteq A^* \cap B^*$ it follows that $ker \; \pi = A^* \cap B^*$.

I suspect that I need to assume that $|M(A \times B)| > |M(A)||M(B)|$ and derive a contradiction. But I'm not sure how to proceed. Any help or comments would be appreciated.

Answer 1

Let $G$ be a Schur cover of $A \times B$, with $G/Z \cong A \times B$ and $Z \le Z(G) \cap [G,G]$ (so $Z$ is $\ker \pi$ in your notation), and let $A^*$ and $B^*$ be the complete inverse images of $A$ and $B$ in $G$. So $G = A^*B^*$ and $A^* \cap B^* = Z$.

Now $[G,G] = [A^*,A^*][B^*,B^*][A^*,B^*]$. We claim that $[A^*,B^*]=1$. To see this note first that $[A^*,B^*] \le A^* \cap B^* = Z$. Now let $a \in A^*$ and $b \in B^*$, and let $m$ and $n$ be the orders of the images of $a$ and $b$ in $A$ and $B$, respectively. Then because elements in $[A^*,B^*]$ are central, we have $[a,b]^m = [a^m,b] = 1 = [a,b^n] = [a,b]^n$ and hence, since $m$ and $n$ are coprime, $[a,b]=1$, proving the claim.

So $[G,G] = [A^*,A^*][B^*,B^*]$ and hence, using $A^* \cap B^* \le Z$, we have $Z = ([A^*,A^*]\cap Z)([B^*,B^*]\cap Z)$. In fact, by a similar argument to the previous paragraph, we can prove that $[A^*,A^*] \cap [B^*,B^*] = 1$, so we have $$Z = ([A^*,A^*]\cap Z) \times ([B^*,B^*]\cap Z).$$

So $[A^*,A^*] \cap Z$ is isomorphic to a quotient group of $M(A)$, and $[B^*,B^*]\cap Z$ is isomorphic to a quotient group of $M(B)$, and the result follows.

Answer 2

If $\Gamma/Z\cong G$ and $Z\leq \mathbf{Z}(\Gamma)\cap \Gamma'$, then $\Gamma$ is called a stem extension of $G$ by $Z$.

(a) Let $\Gamma_A$ and $\Gamma_B$ be stem extension of $G$ by $M(A)$ and $M(B)$, respectively. Because $(\Gamma_A\times\Gamma_B)/(M(A)\times M(B))\cong A\times B$ and $M(A)\times M(B)\leq \mathbf{Z}(A\times B)\cap (A\times B)'$, $|M(A\times B)|\geq|M(A)\times M(B)|$.

(b) Let $\Gamma$ be a stem extension of $A\times B$ by $Z=M(A\times B)$. There is $S, T\lhd \Gamma$ such that $S/Z=A$ and $T/Z=B$. $\Gamma=ST$ and $S\cap T=Z$. Let $\pi_A$ and $\pi_B$ be the set of prime divisors of $|A|$ and $|B|$, respectively. Let $Z_A=\mathbf{O}_{\pi_A}(Z)$ and $Z_B=\mathbf{O}_{\pi_B}(Z)$, then $Z=Z_A\times Z_B$ because every prime divisor of $|Z|$ is contained in $\pi_A$ or $\pi_B$ (by Corollary 5.4), $Z$ is abelian and $\pi_A\cap \pi_B=\varnothing$. Because $|S|=|A||Z_A||Z_B|$ and $|T|=|B||Z_B||Z_A|$, $\gcd(|Z_B|,|S:Z_B|)=\gcd(|Z_A|,|T:Z_A|)=1$, so there are subgroup $X$ and $Y$ such that $S=Z_B X$, $T=Z_A Y$, $|X|=|A||Z_A|$ and $|Y|=|B||Z_B|$ by Schur-Zassenhaus theorem. For all $g\in \Gamma$, $S=S^g={Z_B}^g X^g=Z_B X^g$ and $T=T^g={Z_A}^g Y^g=Z_A Y^g$, there is $s\in Z_B$, $t\in Z_A$ such that $X^g=X^s$, $Y^g=Y^t$ by conjugation part of Schur-Zassenhaus theorem ($Z_A$ and $Z_B$ are abelian). Because $s, t\in Z\leq \mathbf{Z}(\Gamma)$, $X^g=X$ and $Y^g=Y$ for all $g\in \Gamma$. So, $X, Y\lhd \Gamma$. Because $|X||Y|=|\Gamma|$ and $\gcd(|X|,|Y|)=1$, $\Gamma=X\times Y$. Because $X=\mathbf{O}_{\pi_A}(\Gamma)$ and $Y=\mathbf{O}_{\pi_B}(\Gamma)$, $Z_A\leq X$ and $Z_B\leq Y$. By Dedekind lemma, $X/Z_A\cong XZ_B/Z_AZ_B=S/Z=A$ and $Y/Z_B\cong Z_A Y/Z_AZ_B=T/Z=B$. Because $G'=X'\times Y'$, $Z_A\leq G'\cap X=X'$ and $Z_B\leq G'\cap Y=Y'$. Also, $Z_A$ and $Z_B$ centralize $X$ and $Y$, respectively. Therefore, $Z_A\leq \mathbf{Z}(X)\cap X'$ and $Z_B\leq \mathbf{Z}(Y)\cap Y'$. Now we know that $\Gamma=X\times Y$, $Z=Z_A\times Z_B$, $Z_A\leq X$ and $Z_B\leq Y$. Also, $X$ and $Y$ are stem extension of $A$ and $B$ by $Z_A$ and $Z_B$, respectively. Because $|M(A)||M(B)|\leq |M(A\times B)|=|Z|=|Z_A||Z_B|\leq |M(A)||M(B)|$, $Z_A\cong M(A)$ and $Z_B\cong M(B)$. Therefore, $M(A\times B)=Z=Z_A\times Z_B\cong M(A)\times M(B)$.