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Problem 83 (The Right And Left Cosets Of The Kernel Are Equal)

Suppose that $f:G\to H$ is a homomorphism. Let $K$ be the kernel of $f$ and let $a\in G$. Prove the following:

  1. The right coset $Ka$ equals the set of values $b\in G$ such that $f(b)=f(a)$, or symbolically we have $$Ka = \{b\in G\mid f(b)=f(a)\} = f^{-1}(f(a)).$$
  2. The left coset $aK$ equals the same set $\{b\in G\mid f(b)=f(a)\}$, which means the left and right cosets are the same or $Ka=aK$.

Solution

1. We first show $Ka \subseteq \{b\in G\mid f(b)=f(a)\}$. Let $x\in Ka$. This means $x=ka$ for some $k\in K$. We compute $f(x)=f(ka)=f(k)f(a)=e_Hf(a)=f(a)$. Hence $x\in \{b\in G\mid f(b)=f(a)\}$. We now show $\{b\in G\mid f(b)=f(a)\} \subseteq Ka$. Let $y\in \{b\in G\mid f(b)=f(a)\}$. Hence $f(y)=f(a)$. Note $y=ye_H=y(a^{-1}a)=(ya^{-1})a$. We now show $ya^{-1}\in K$. We compute

$$ \begin{align} f(ya^{-1})&=f(y)f(a^{-1})\\ &=f(a)f(a^{-1})\\ &=f(aa^{-1})\\ &=f(e_G)\\ &=e_H \end{align} $$

Since $ya^{-1}\in K$, and $y=(ya^{-1})a$, we conclude that $y\in Ka$.

2. We first show $aK \subseteq \{b\in G\mid f(b)=f(a)\}$. Let $x\in aK$. This means $x=ak$ for some $k\in K$. We compute $f(x)=f(ak)=f(a)f(k)=f(a)e_H=f(a)$. Hence $x\in \{b\in G\mid f(b)=f(a)\}$. We now show $\{b\in G\mid f(b)=f(a)\} \subseteq aK$. Let $y\in \{b\in G\mid f(b)=f(a)\}$. Hence $f(y)=f(a)$. Note $y=e_Hy=(aa^{-1})y=a(a^{-1}y)$. We now show $a^{-1}y\in K$. We compute

$$ \begin{align} f(a^{-1}y)&=f(a^{-1})f(y)\\ &=f(a^{-1})f(a)\\ &=f(a^{-1}a)\\ &=f(e_G)\\ &=e_H \end{align} $$

Since $a^{-1}y\in K$, and $y=a(a^{-1}y)$, we conclude that $y\in aK$.

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