Exercise 3.1.4 ($(gN)^\alpha = g^\alpha N$ for all $\alpha \in \mathbb{Z}$)

Question

Prove that in the quotient group $G/N$, $(gN)^\alpha = g^\alpha N$ for all $\alpha in \mathbb{Z}$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

\begin{proof} Consider for every $\alpha \in \Z$ the proposition
$$\mathscr{P}(\alpha) \iff (gN)^\alpha = g^\alpha N.$$
First $(gN)^0 = 1_{G/N} = N$ and $g^0 N = 1_G \cdot N = N$, so $\mathscr{P}(0)$ is true.

Suppose now that $\mathscr{P}(\alpha)$ is true for some integer $\alpha \geq 0$, so that $(gN)^\alpha = g^\alpha N$. By definition of the law in $G/N$,
\begin{align*}
(gN)^{\alpha + 1} &= (gN)^\alpha gN\
&= g^\alpha N \, gN\
&= g^\alpha g N\
&= g^{\alpha + 1} N.
\end{align*}
so $\mathscr{P}(\alpha + 1)$ is true.

The induction is done, which proves that
$$\forall \alpha \in \N,\ (gN)^\alpha = g^\alpha N.$$
We know that $(gN)^{-1} = g^{-1} N$. Therefore, for $\alpha \geq 0$,
\begin{align*}
(gN)^{-\alpha} &= \left((gN)^\alphaight)^{-1}\
&= \left(g^{\alpha} Night)^{-1}\
&= (g^\alpha)^{-1} N\
&= g^{-\alpha} N.
\end{align*}
This shows that
$$\forall \alpha \in \Z,\ (gN)^\alpha = g^\alpha N.$$
\end{proof}

Exercise 3.1.4 ($(gN)^\alpha = g^\alpha N$ for all $\alpha \in \mathbb{Z}$)

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Exercise 3.1.4 ($(gN)^\alpha = g^\alpha N$ for all $\alpha \in \mathbb{Z}$)

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