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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

Recall, the definition of the kernel of an homomorphism. Also, let $S = \{b\in G\mid f(b)=f(a)\}$.

Part 1: First, we will show that for all $b \in Ka$, it is true that $b \in S$, and then we will show that for all $c \in S$, it is true that $c \in Ka$. Let $b \in Ka$. Let $k \in K$, such that $b = ka$. We compute $$ f(b) = f(ka) = f(k)f(a) = e_Hf(a) = f(a) $$ Thus, we know that $b \in S$. Let $c \in S$. Thus, we know that $c = (ca^{-1}a)$. We now show that $ca^{-1} \in K$. We compute $$f(ca^{-1}) = f(c)f(a^{-1}) = f(c)f(a)^{-1} = f(a)f(a)^{-1} = e_H.$$ Thus, we know that $c \in Ka$.

Part 2: First, we will show that for all $b \in aK$, it is true that $b \in S$, and then we will show that for all $c \in S$, that $c \in aK$. Let $b \in aK$. Let $k \in K$, such that $b = ak$. We compute $$ f(b) = f(ak) = f(a)f(k) = f(a)e_H = f(a) $$ Thus, we know that $b \in S$. Let $c \in S$. Thus, we know that $c = aa^{-1}c \in K$. We now show that $a^{-1}c \in K$. We compute $$f(a^{-1}c) = f(a^{-1})f(c) = f(a)^{-1}f(c) = f(a)^{-1}f(a) = e_H.$$ Thus, we know that $c \in aK$. Thus, we know that $Ka = aK$.

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