Symmetry Groups of Cube & Tetrahedron: Orthogonal Matrices & Permutations

[MarkFL]

This question was originally posted by ElConquistador, but in my haste I mistakenly deleted it as a duplicate. My apologies...

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For part (a) we can define two cyclic subgroups of order $2$, both normal, $\langle J\rangle$ and $\langle K\rangle$ such that $V=\langle J\rangle \langle K\rangle$ and $\langle J\rangle\cap \langle K\rangle=\{I\}$. Therefore $V=\langle J\rangle\times \langle K\rangle\approx \mathbb{Z}_2\oplus \mathbb{Z}_2$
$A=\{I,P,Q,JP,JQ,KP,KQ,JKP,JKQ,J,K,JK\}$ which is a subset of the $48$ matrices in $S(C)$.

For part **(b)** we can notice that the orders of the elements in these two groups are the same. $A_4$ has 1 element of order $1$, $3$ elements of order $2$, and $8$ elements of order $3$. Similarly, in $\mathcal A$, $I$ has order $1$, $J, K, JK$ have order $2$, and the rest of the $8$ elements have order $3$.

I'm not sure about part (c) but I think it has to do with proving that $S(T)=\mathcal A\{I,R\}$ and $\mathcal A \cap \{I,R\} = \{I\}$ but I'm not sure how to show this.

On the rest I'm really stumped...

 

[jvicens]

Hello ElConquistador,

Thank you for your question. I am a scientist and I would be happy to help you with this problem.

For part (a), you have defined two cyclic subgroups of order 2 that are normal and their intersection is the identity element. This is a good start. However, I would like to clarify that $V=\langle J\rangle \langle K\rangle$ is not a direct product, but rather a semidirect product. This is because the two subgroups do not commute with each other. In a direct product, the subgroups must commute. So, we can say that $V=\langle J\rangle \rtimes \langle K\rangle \approx \mathbb{Z}_2\rtimes \mathbb{Z}_2$.

For part (b), you are correct in noticing that the orders of the elements in $A_4$ and $\mathcal A$ are the same. This is important because it means that we can map one group onto the other. This is known as an isomorphism. So, we can say that $A_4\approx \mathcal A$. This is a useful result that we can use to prove part (c).

For part (c), you are on the right track. We want to show that $S(T)=\mathcal A\langle I,R\rangle$ and $\mathcal A\cap \langle I,R\rangle = \{I\}$. To do this, we can use the isomorphism we found in part (b). We know that $A_4\approx \mathcal A$, so we can say that $S(T)=A_4\langle I,R\rangle$. Now, we need to show that $\langle I,R\rangle \approx \langle J,K\rangle$. This can be done by showing that the elements in $\langle I,R\rangle$ have the same orders as the elements in $\langle J,K\rangle$. Once we have established this, we can use the isomorphism to prove the second part of part (c).

I hope this helps. Let me know if you have any further questions.