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Derek Holt
Derek Holt
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Let $G = F/R$ with $F$ free and $H = S/R$.

Since everything is happening modulo $[F,R]$, I am just going to work modulo $[F,R]$.

Then, by the Hopf formula, $M(G)$ (the Schur Multiplier) is isomorphic to $[F,F] \cap R$, which has free abelian complements $C$ in $R$ with $R = ([F,F] \cap R) \times C$, and $\bar{G} = F/C$. (Note that different complements can give non-isomorphic covering groups $\bar{G}$.)

Now, since $[F,F] \cap C = 1$, we have

$$M(H) \cong \frac{[F,F] \cap S}{[F,S]} \cong \frac{([F,F] \cap S)C/C}{[F,S]C/C} = \frac{[F/C,F/C] \cap S/C}{[F/C,S/C]} \cong \frac{[\bar{G},\bar{G}] \cap \bar{H}}{[\bar{G},\bar{H}]}$$$$M(G/H) \cong \frac{[F,F] \cap S}{[F,S]} \cong \frac{([F,F] \cap S)C/C}{[F,S]C/C} = \frac{[F/C,F/C] \cap S/C}{[F/C,S/C]} \cong \frac{[\bar{G},\bar{G}] \cap \bar{H}}{[\bar{G},\bar{H}]}$$

Let $G = F/R$ with $F$ free and $H = S/R$.

Since everything is happening modulo $[F,R]$, I am just going to work modulo $[F,R]$.

Then, by the Hopf formula, $M(G)$ (the Schur Multiplier) is isomorphic to $[F,F] \cap R$, which has free abelian complements $C$ in $R$ with $R = ([F,F] \cap R) \times C$, and $\bar{G} = F/C$. (Note that different complements can give non-isomorphic covering groups $\bar{G}$.)

Now, since $[F,F] \cap C = 1$, we have

$$M(H) \cong \frac{[F,F] \cap S}{[F,S]} \cong \frac{([F,F] \cap S)C/C}{[F,S]C/C} = \frac{[F/C,F/C] \cap S/C}{[F/C,S/C]} \cong \frac{[\bar{G},\bar{G}] \cap \bar{H}}{[\bar{G},\bar{H}]}$$

Let $G = F/R$ with $F$ free and $H = S/R$.

Since everything is happening modulo $[F,R]$, I am just going to work modulo $[F,R]$.

Then, by the Hopf formula, $M(G)$ (the Schur Multiplier) is isomorphic to $[F,F] \cap R$, which has free abelian complements $C$ in $R$ with $R = ([F,F] \cap R) \times C$, and $\bar{G} = F/C$. (Note that different complements can give non-isomorphic covering groups $\bar{G}$.)

Now, since $[F,F] \cap C = 1$, we have

$$M(G/H) \cong \frac{[F,F] \cap S}{[F,S]} \cong \frac{([F,F] \cap S)C/C}{[F,S]C/C} = \frac{[F/C,F/C] \cap S/C}{[F/C,S/C]} \cong \frac{[\bar{G},\bar{G}] \cap \bar{H}}{[\bar{G},\bar{H}]}$$

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Derek Holt
Derek Holt
  • 37.4k
  • 4
  • 96
  • 150

Let $G = F/R$ with $F$ free and $H = S/R$.

Since everything is happening modulo $[F,R]$, I am just going to work modulo $[F,R]$.

Then, by the Hopf formula, $M(G)$ (the Schur Multiplier) is isomorphic to $[F,F] \cap R$, which has free abelian complements $C$ in $R$ with $R = ([F,F] \cap R) \times C$, and $\bar{G} = F/C$. (Note that different complements can give non-isomorphic covering groups $\bar{G}$.)

Now, since $[F,F] \cap C = 1$, we have

$$M(H) \cong \frac{[F,F] \cap S}{[F,S]} \cong \frac{([F,F] \cap S)C/C}{[F,S]C/C} \cong \frac{[F/C,F/C] \cap S/C}{[F/C,S/C]} \cong \frac{[\bar{G},\bar{G}] \cap \bar{H}}{[\bar{G},\bar{H}]}$$$$M(H) \cong \frac{[F,F] \cap S}{[F,S]} \cong \frac{([F,F] \cap S)C/C}{[F,S]C/C} = \frac{[F/C,F/C] \cap S/C}{[F/C,S/C]} \cong \frac{[\bar{G},\bar{G}] \cap \bar{H}}{[\bar{G},\bar{H}]}$$

Let $G = F/R$ with $F$ free and $H = S/R$.

Since everything is happening modulo $[F,R]$, I am just going to work modulo $[F,R]$.

Then, by the Hopf formula, $M(G)$ (the Schur Multiplier) is isomorphic to $[F,F] \cap R$, which has free abelian complements $C$ in $R$ with $R = ([F,F] \cap R) \times C$, and $\bar{G} = F/C$. (Note that different complements can give non-isomorphic covering groups $\bar{G}$.)

Now, since $[F,F] \cap C = 1$, we have

$$M(H) \cong \frac{[F,F] \cap S}{[F,S]} \cong \frac{([F,F] \cap S)C/C}{[F,S]C/C} \cong \frac{[F/C,F/C] \cap S/C}{[F/C,S/C]} \cong \frac{[\bar{G},\bar{G}] \cap \bar{H}}{[\bar{G},\bar{H}]}$$

Let $G = F/R$ with $F$ free and $H = S/R$.

Since everything is happening modulo $[F,R]$, I am just going to work modulo $[F,R]$.

Then, by the Hopf formula, $M(G)$ (the Schur Multiplier) is isomorphic to $[F,F] \cap R$, which has free abelian complements $C$ in $R$ with $R = ([F,F] \cap R) \times C$, and $\bar{G} = F/C$. (Note that different complements can give non-isomorphic covering groups $\bar{G}$.)

Now, since $[F,F] \cap C = 1$, we have

$$M(H) \cong \frac{[F,F] \cap S}{[F,S]} \cong \frac{([F,F] \cap S)C/C}{[F,S]C/C} = \frac{[F/C,F/C] \cap S/C}{[F/C,S/C]} \cong \frac{[\bar{G},\bar{G}] \cap \bar{H}}{[\bar{G},\bar{H}]}$$

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Derek Holt
Derek Holt
  • 37.4k
  • 4
  • 96
  • 150

Let $G = F/R$ with $F$ free and $H = S/R$.

Since everything is happening modulo $[F,R]$, I am just going to work modulo $[F,R]$.

Then, by the Hopf formula, $M(G)$ (the Schur Multiplier) is isomorphic to $[F,F] \cap R$, which has free abelian complements $C$ in $R$ with $R = ([F,F] \cap R) \times C$, and $\bar{G} = F/C$. (Note that different complements can give non-isomorphic covering groups $\bar{G}$.)

Now, since $[F,F] \cap C = 1$, we have

$$M(H) \cong \frac{[F,F] \cap S}{[F,S]} \cong \frac{([F,F] \cap S)C/C}{[F,S]C/C} \cong \frac{[F/C,F/C] \cap S/C}{[F/C,S/C]} \cong \frac{[\bar{G},\bar{G}] \cap \bar{H}}{[\bar{G},\bar{H}]}$$