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Given the matrices with coefficients in $\mathbb Z_5$, I am asked to consider: $G= \left\{\begin{pmatrix} a & 0 \\ 0 &d \end{pmatrix} \mid ad\neq 0 \bmod 5 \right\}$ and: $ H= \left\langle \begin{pmatrix} 1 & 0 \\ 0 &-1 \end{pmatrix}, \begin{pmatrix} -1 & 0 \\ 0 &1 \end{pmatrix}\right \rangle$

And find matrices $A_1, A_2, A_3, A_4$ in $G$, such that: $$ G/H= \{A_1H, A_2 H, A_3 H, A_4 H \} $$

I know that a standard coset is definitely the set $H$ itself, so a good idea would be to consider: $A_k=kI$ for $k=1,2,3,4$ How can I efficiently check that this works? (or not) I mean, I can compute the action of these matrices on the 4 elements of $H$ and show it constructs all 16 elements each in their own little 4-sized coset, but is there a faster way?


Edit: This partitions does not seem to work:

$$H= \begin{pmatrix} 1 & 0 \\ 0 &-1 \end{pmatrix} ,\begin{pmatrix} -1 & 0 \\ 0 &1 \end{pmatrix} , \begin{pmatrix} -1 & 0 \\ 0 &-1 \end{pmatrix} , \begin{pmatrix} 1 & 0 \\ 0 &1 \end{pmatrix} $$ We will now compute all these matrices modulo 5 and check whether or not we partition all matrices into their own coset. We should get 4 disjoint sets that represent the equivalence classes of $G/H$. Notice that matrix multiplication by a diagonal matrix of the form $\lambda I$ just scales all the entries by $\lambda$. $H= \begin{pmatrix} 1 & 0 \\ 0 &4 \end{pmatrix} ,\begin{pmatrix} 4 & 0 \\ 0 &1 \end{pmatrix} , \begin{pmatrix} 4 & 0 \\ 0 &4 \end{pmatrix} , \begin{pmatrix} 1 & 0 \\ 0 &1 \end{pmatrix} $ $ 2H= \begin{pmatrix} 2 & 0 \\ 0 &8 \end{pmatrix} ,\begin{pmatrix} 8 & 0 \\ 0 &2 \end{pmatrix} , \begin{pmatrix} 8 & 0 \\ 0 &8 \end{pmatrix} , \begin{pmatrix} 2 & 0 \\ 0 &2 \end{pmatrix} \equiv \begin{pmatrix} 2 & 0 \\ 0 &3 \end{pmatrix} ,\begin{pmatrix} 3 & 0 \\ 0 &2 \end{pmatrix} , \begin{pmatrix} 3 & 0 \\ 0 &3 \end{pmatrix} , \begin{pmatrix} 2 & 0 \\ 0 &2 \end{pmatrix} \bmod 5 $ $3H= \begin{pmatrix} 3 & 0 \\ 0 & 12 \end{pmatrix} ,\begin{pmatrix} 12 & 0 \\ 0 &3 \end{pmatrix} , \begin{pmatrix} 12 & 0 \\ 0 &12 \end{pmatrix} , \begin{pmatrix} 3 & 0 \\ 0 &3 \end{pmatrix} \equiv \begin{pmatrix} 3 & 0 \\ 0 &2 \end{pmatrix} ,\begin{pmatrix} 3 & 0 \\ 0 &2 \end{pmatrix} , \begin{pmatrix} 2 & 0 \\ 0 &2 \end{pmatrix} , \begin{pmatrix} 3 & 0 \\ 0 &3 \end{pmatrix} \bmod 5 $ $4H= \begin{pmatrix} 4 & 0 \\ 0 & 16 \end{pmatrix} ,\begin{pmatrix} 16 & 0 \\ 0 &4 \end{pmatrix} , \begin{pmatrix} 16 & 0 \\ 0 &16 \end{pmatrix} , \begin{pmatrix} 4 & 0 \\ 0 &4 \end{pmatrix} \equiv \begin{pmatrix} 4 & 0 \\ 0 &1 \end{pmatrix} ,\begin{pmatrix} 1 & 0 \\ 0 &4 \end{pmatrix} , \begin{pmatrix} 1 & 0 \\ 0 &1 \end{pmatrix} , \begin{pmatrix} 4 & 0 \\ 0 &4 \end{pmatrix} \bmod 5 $

So close! but it does not work =(

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  • $\begingroup$ See also this question. $\endgroup$ – Dietrich Burde Feb 26 at 16:23
  • $\begingroup$ Your $A_1 = I$ is in $H$, because $\begin{pmatrix} 1 & 0 \\ 0 &-1 \end{pmatrix}\cdot \begin{pmatrix} 1 & 0 \\ 0 &-1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 &1 \end{pmatrix}$. So this means that $A_1$ will not generate a coset apart from the trivial one. $\endgroup$ – Snake707 Feb 26 at 16:25
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    $\begingroup$ @Snake707 That in itself is not a problem; if you want all the cosets, you need to make sure one of the ones you have is $H$. $\endgroup$ – Morgan Rodgers Feb 26 at 16:35
  • $\begingroup$ example: $G=Z_8$ under addition, we know $H=\{0,4\}$ is a subgroup, now the cosets of G/H can be given by $0+H$, $1+H$, $2+H$, $3+H$, again $0$ is in $H$, not a problem though. $\endgroup$ – Wesley Strik Feb 26 at 16:42
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    $\begingroup$ Well $H$ has $4$ elements and $G$ has $16$. You are proposing 4 cosets. If one of those cosets is identical to another, this will make it a problem, especially since you also noticed that $A_4H = A_1H = H$. So you need additional cosets. $\endgroup$ – Snake707 Feb 26 at 17:27
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Actually, garnering from your answer, the algorithm for this specific problem is quite simple. Let $H$ be a subgroup of $G$, then it is known for two cosets $aH$, $bH$ it is either $aH = bH$ or $aH \cap bH = \emptyset$.

So you start by looking at what elements $G$ has. Then $H$ is your first coset. Take any element $A_1 \in G - H$. Then $A_1H$ is your next coset. Take any $A_2 \in G - (H \cup A_1H)$ and then $A_2H$ is your next coset. Continue until $G$ is completely partitioned.

I suggest you try out $A_3 = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$ and $A_4 = \begin{pmatrix} 2 & 0 \\ 0 &1 \end{pmatrix}$.

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  • $\begingroup$ I will give it a try, check back with you later. $\endgroup$ – Wesley Strik Feb 26 at 17:40
  • $\begingroup$ Yes, and it is actually a good general approach, thank you! $\endgroup$ – Wesley Strik Feb 27 at 9:24

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