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In my introductory abstract algebra course, the quotient group $G/H$ was defined as $$G/H=\{gH:g\in G\}$$ which is a set of sets. In an exercise, I should show that for the group of invertible matrices $GL_n(K)$ over a field $K$ and the normal subgroup $SL_n(K)$ the quotient group is abelian.

I'm horribly confused. What is the operation that combines two sets of matrices? What does it mean for two sets of matrices to commute with respect to this operation?

I apologize if this is a silly question, but our lecture only ever mentioned modular arithmetic…

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2 Answers 2

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The application $\text{det} : \text{GL}_{n}(K) \rightarrow K^{*}$ is a homomorphism and his kernel is $\text{SL}_{n}(K)$. Moreover $\text{det}$ is surjective. Then $\frac{\text{GL}_{n}(K)}{\text{SL}_{n}(K)}$ is isomorphic to the commutative group $K^{*}$.

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  • $\begingroup$ Thanks for the typo. $\endgroup$
    – CechMS
    Commented May 7, 2020 at 10:17
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The operation in $G/H$ is: $(g_1H).(g_2H)=(g_1g_2)H$. And $G/H$ is commutative if and only if, for any $g_1,g_2\in G$, $(g_1g_2)H=(g_2g_1)H$. But\begin{align}(g_1g_2)H=(g_2g_1)H&\iff(g_1g_2)^{-1}(g_1g_2)\in H\\&\iff g_1^{\,-1}g_2^{\,-1}g_1g_2\in H.\end{align}In the case in which $G=GL_n(k)$ and $H=SL_n(k)$, this is true, because$$\det(g_1^{\,-1}g_2^{\,-1}g_1g_2)=\det(g_1)\det(g_2)\det(g_1)^{-1}\det(g_2)^{-1}=1$$and therefore $g_1^{\,-1}g_2^{\,-1}g_1g_2\in SL_n(k)$.

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  • $\begingroup$ Could you please expand on why $(g_1g_2)(g_1g_2)^{-1}$ must be in $H$? $\endgroup$ Commented May 7, 2020 at 9:04
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    $\begingroup$ Because $g_1H=g_2H$ means that $g_1g_2^{-1}\in H$. $\endgroup$ Commented May 7, 2020 at 9:21

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