If $H$ is a subgroup of $G$ prove that the set $G/H$ of left cosets is a group with product $(aH)(bH) = (abH)$ if and only if $H$ is a normal subgroup of $G$.
attempt: Suppose coset multiplication is well defined.
Let $g\in G$, and $h\in H$. THen $1*H = hH$. Then we have $g^{-1}H = (1*H)(g^{-1}H) = (hH)(g^{-1}H) = (hg^{-1}H)$ so that $hg^{-1} \in g^{-1}H$ and $ghg^{-1} \in H$.
Conversely,suppose H is a normal subgroup of G. Then
we wills how the set $G/H$ of left cosets is a group with product $(aH)(bH) = (abH)$.
1)associative: (aH * bH)cH = (ab)H * cH = (ab)cH
and aH(bH * cH) = aH* (bc)H = a(bc)H.
2)Identity: 1H*gH = gH = gH * 1H for all $g\in G$
thus, $H = 1*H$ is the identity for coset multiplication.
3) inverse: $(gH)(g^{-1}H) = (1H) = g^{-1}*gH = 1*H$ for all $g \in G$
thus $(gH)^{-1} = g^{-1}H$
Can someone please verify this is correct? Any feedback or better way of approaching this would really be appreciated it.
Thank you.
Best Answer
Suppose $H$ is a normal subgroup of $G$. Then $ghg^{-1} \in H$ for all $h \in H$.
Choosing coset representatives, $h_1$, $h_2$, using the normality of $H$ and the closure of $H$ under multiplication ($h_ih_j \in H$), we can write $$(ah_1)(bh_2) = ab(b^{-1}h_1b)h_2 = ab(h_3h_2) =ab(h_4).$$ This shows that $(aH)(bH) = (ab)H$. With a well-defined operation on the cosets you can proceed as above to confirm associativity, identity, and inverses.