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**Statement:**

Let $H$ be a subgroup of a group $G$.Then left coset multiplication is well defined by the equation

(1)if and only if $H$ is a normal subgroup of $G$.

**Proof:**

Suppose first that $(aH)(bH)=(ab)H$ does give a well-defined binary operation on left cosets. Let $a\in G$. We want to show that $aH$ and $Ha$ are the same set. We use the standard technique of showing that each is a subset of the other.

Let $x\in aH$. Choosing representatives $x\in aH$ and $a^{-1}\in a^{-1}H$, we have $(xH)(a^{-1}H)=(xa^{-1})H$. On the other hand, choosing representatives $a\in aH$ and $a^{-1}\in a^{-1}H$, we see that $(aH)(a^{-1}H)=eH=H$. Using our assumption that left coset multiplication by representatives is well defined, we must have $xa^{-1}=h\in H$. Then $x=ha$, so $x\in Ha$ and $aH\subseteq Ha$.

We turn now to the converse: If $H$ is a normal subgroup, then left coset multiplication by representatives is well-defined. Due to our hypothesis, we can simply say *cosets*, omitting *left* and *right*. Suppose we wish to compute $(aH)(bH)$. Choosing $a\in aH$ and $b\in bH$, we obtain the coset $(ab)H$. Choosing different representatives $ah_1\in aH$ and $bh_2\in bH$, we obtain the coset $ah_1bh_2H$. We must show that these are the same cosets. Now $h_1b\in Hb=bH$, so $h_1b=bh_3$ for some $h_3\in H$. Thus

and $(ab)(h_3h_2)\in (ab)H$. Therefore, $ah_1bh_2$ is in $(ab)H$.