Theorem 13.15

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Let $\phi: G\rightarrow G'$ be a group homomorphism, and let $H=Ker(\phi)$.Let $a\in G$.Then the set

\begin{align} \phi^{-1}[\{\phi(a)\}]=\{x\in G|\phi(x)=\phi(a)\} \end{align}

is the left coset $aH$ of $H$,and is also the right coset $Ha$ of $H$.Consequently,the two partitions of $G$ into left cosets and into right cosets of $H$ are the same.

We want to show that

\begin{align} \{x\in G|\phi(x)=\phi(a)\}=aH. \end{align}

(There is a standard way to show that two sets are equal;show that each is a subset of the other.)
Suppose that $\phi(x)=\phi(a)$.Then

\begin{align} \phi(a)^{-1}\phi(x)=e'. \end{align}

where $e'$ is the identity of $G'$. By Theorem 13.12, we know that $\phi(a)^{-1}=\phi(a^{-1})$,so we have

\begin{align} \phi(a)^{-1}\phi(x)=e'. \end{align}

Since $\phi$ is a homomorphism,we have

\begin{align} \phi(a)^{-1}\phi(x)=\phi(a^{-1}x), \end{align}

so $\phi(a^{-1}x)=e'$.
But this shows that $a^{-1}x$ is in $H=Ker(\phi)$,so $a^{-1}x=h$ for some $h\in H$,and $x=ah\in aH$.This shows that

\begin{align} \{x\in G|\phi(x)=\phi(a)\}\subseteq aH. \end{align}

To show containment in the other direction,let $y\in aH$,so that $y=ah$ for some $h\in H$.Then

\begin{align} \phi(y)=\phi(ah)=\phi(a)\phi(h)=\phi(a)e'=\phi(a), \end{align}

so that $y\in\{x\in G|\phi(x)=\phi(a)\}$.

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