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$G$ is $S_3$.

$H = \{I,A,B,C,D,K\}$ where $$ I= \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \quad A=\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} \quad B = \begin{pmatrix} 0 & 1 \\ -1 & -1 \\ \end{pmatrix} $$

$$ C= \begin{pmatrix} -1 & -1 \\ 0 & 1 \\ \end{pmatrix} \quad D=\begin{pmatrix} -1 & -1 \\ 1 & 0 \\ \end{pmatrix} \quad K=\begin{pmatrix} 1 & 0 \\ -1 & -1 \\ \end{pmatrix} $$ and where H is a group under matrix multiplication. Determine whether or not G is isomorphic to H.

Is there a good way to go about this besides writing out their group tables?

I know to show a group is isomorphic to another group is to show that there is an isomorphism, i.e., that there is a bijective function $f:G \rightarrow H$ such that $f(ab)=f(a)f(b), \text{with} \ a,b \in G.$ I'm new with isomorphisms and don't really understand how to see if a group is isomorphic to another.

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There are many good ways. Basic methods would be finding a small generating set for one of the two groups and trying to generate a homomorphism from a choice of image of that generating set, comparing the order of the two groups, and comparing the subgroup structure of the two groups. – Loki Clock Mar 14 '14 at 22:42

Another way, if you don't want to write tables,… and if you know that up to isomorphism there are only 2 groups of order $6$, is to note that $BC\neq CB$ so $H$ has to be isomorphic to $S_3$ since it can't be isomorphic to the "other" group of order $6$, i.e. $\mathbb Z/6\mathbb Z$.

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One way to show it is to notice that $H$ operates on $\{(1,1),(-2,1),(1,-2)\}$ by writing out the action of each matrix explicitly.

(This is also a quicker way to show that $H$ is actually a group than doing all of the matrix products, in case you're supposed to show that).

Which hat did I pull that out of? I know that $S_3$ is isomorphic to the dihedral group $D_6$ with 6 elements, i.e. the symmetries of an equilateral triangle. So if $H$ is isomorphic to it, perhaps it has been constructed as those symmetries expressed in some convenient coordinate system.

$A$ has order 2, so it would correspond to a reflection, and indeed it reflects about the line $x=y$. But all the coefficients in $H$ are integers, so it cannot be an ordinary octogonal coordinate system. On the other hand, we'd get integers if we imagine two axes meeting at 60°. Jot down a sketch of that and find a suitable equilateral triangle that has the bisector between the axes as an altitude. The smallest such has corners $(1,1),(-2,1),(1,-2)$.

And then it's just a matter of checking that this guess actually works.

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Why that choice of vectors? – Loki Clock Mar 14 '14 at 22:49

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