Show that the direct product of the group of symmetries of the square and the cyclic group of order 2 has the following presentation.

Question

Show that the direct product of the group of symmetries of the square and the cyclic group of order 2 has the following presentation:

$$G=\langle a,b,c\mid a^2,b^2,(ab)^4,c^2,bc=cb,ac=ca\rangle.$$

I made the Cayley table for this group and for the choices of $a,b,c$ I have 3x2 options. (Since $D4$ has 3 elements of order 2). But it seems impossible to avoid $(ab)^2 = 1 $ which would make the group Abelian which cannot happen.

Use $D_4=\langle a,b\mid a^4=b^2=e, ab=ba^{-1}\rangle$. So definitely $ab\neq ba$, and $G=D_4\times \langle c\rangle$ is not abelian. Of course $C_2=\langle c\mid c^2=e\rangle$. — Dietrich Burde, Sep 01 '24 at 12:49

Shaun · Accepted Answer · 2024-09-01T13:24:50.140

3

Use this standard result about direct products, and the fact that

$$D_{2n}\cong\langle a,b\mid a^2, b^2, (ab)^n\rangle.$$

edited Sep 01 '24 at 13:24

answered Sep 01 '24 at 13:00

Shaun

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Show that the direct product of the group of symmetries of the square and the cyclic group of order 2 has the following presentation.

1 Answers1