Identifying $\langle x_1, x_2, x_3, x_4\mid x_1x_2=x_3, x_3x_2=x_1, x_1x_4=x_3, x_3x_4=x_1\rangle$.

Question

What group is $\langle x_1, x_2, x_3, x_4\mid x_1x_2=x_3, x_3x_2=x_1, x_1x_4=x_3, x_3x_4=x_1\rangle$?

Thoughts:

The group is infinite according to GAP, $x_2=x_4$, and $x_2^2=id.$

Yes, @AlexProvost: use Cayley's theorem to convert $C_2\times C_2$ into a group of permutations, namely ${id., (13), (24), (13)(24)}$, then permute the generators according to these permutations. — Shaun, Nov 02 '17 at 10:56

score 2 · Accepted Answer · answered Nov 02 '17 at 10:56

2

Setting $x_3:=x_1x_2$ the relations become $x_1x_2^2=x_1$, so $x_2^2=1$, and $x_4=x_1^{-1}x_1x_2=x_2$. The last relation $x_3x_4=x_1$ then just says $x_1=x_1$. So, writing $x_1=a,x_2=b$ the group is given by $$ \langle a,b \mid b^2=1\rangle, $$ and now you see it.

answered Nov 02 '17 at 10:56

Dietrich Burde

130,978

So, it's the free product of $\Bbb Z$ with $C_2$? – Shaun Nov 02 '17 at 10:59
Right - in contrast to this group. – Dietrich Burde Nov 02 '17 at 11:58

Identifying $\langle x_1, x_2, x_3, x_4\mid x_1x_2=x_3, x_3x_2=x_1, x_1x_4=x_3, x_3x_4=x_1\rangle$.

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