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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.

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  • $\begingroup$ The proof of the converse doesn't appear to use the normality of $H$ at all; you need to show that the coset multiplication is well-defined to begin with. $\endgroup$ – Titus Sep 16 '15 at 22:00
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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.

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  • $\begingroup$ Thank you! Is the forward direction fine then? $\endgroup$ – Mahidevran Sep 16 '15 at 22:12
  • $\begingroup$ Yeah, it looked good. $\endgroup$ – Titus Sep 16 '15 at 22:13

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