An interesting group

I like algebra. MAT5420 was my first mathematics course in Wayne State. Here’s a problem that gave me entertainment this semester. It’s not from MAT5420. It’s an exercise from Serre’s “Trees”.

Show that the group $G = \langle a, b, c \ | \ bab^{-1} = a^2, \ \ cbc^{-1} = b^2, \ \ aca^{-1} = c^2 \rangle$ is trivial.

Here is one way to show it. It is just a brute force computation. The three relations can be written as follows:

$bab^{-1} = a^2 \text{ as } ba=a^2b,$ $cbc^{-1} = b^2 \text{ as } cb=b^2c,$ and $aca^{-1} = c^2 \text{ as } ac=c^2a.$
We just need to manipulate them in the right way to get the final result.

First,
$a^4cb= (bab^{-1})(bab^{-1})(cb)$
$= (ba^2b^{-1})(b^2c)$
$= ba^2bc$
$= b(ba)c$
$= b^2ac$
$= b^2c^2 a.$

Second, since $a^4c= a^3 a c$
$a^4c= a^3c^2a$
$= a^2(ac)ca$
$= a^2(c^2a)ca$
$= a(ac)(c)(ac)a$
$= a(c^2a)(c)(c^2a)a$
$= (ac)c(ac)ccaa$
$= (c^2a)c(c^2a)cca^2$
$= c^2c^2(ac)c(ac)ca^2$
$= c^4(c^2a)(c)(c^2a)ca^2$
$= c^6(ac)(c^2)(ac)a^2$
$= c^6(c^2a)(c^2)(c^2a)(a^2)$
$= c^8 (ac)(c)(c^2a)(a^2)$
$= c^8(c^2a)(c)(c^2)a^3$
$= c^{10}(ac)(c^2)a^3$
$= c^{10}c^2(ac)(c)(a^3)$
$= c^{12}(ac)(c)(a^3)$
$= c^{12}(c^2a)(c)(a^3)$

$= c^{14}(ac)(a^3)$

$= c^{14}(c^2a)(a^3)$
$= c^{16}a^4$

we have $a^4cb=c^{16}{a^4}{b}$ . Since $a^4b=ba^2$ , we have
$a^4cb =c^{16}{a^4}{b} = c^{16}ba^2$ .
Now, rather than do the above long calculation all over again to see that

$c^{16}b = b^{2^{16}}c^{16}$ ,

we can use induction. Our base step says $cb=b^2c$ . Our induction step is $c^kb=b^{2^k}c^k$ . Then

$latex c^{k+1}b = c^k(cb)
$= c^k(b^2c)$
$= c^k(b)(bc)$
$= b^{2^k}c^k bc$
$= b^{2^k}(c^k b)c$
$= b^{2^k}(b^{2^k}c^{k}c$
$= b^{2^k}(b^{2^k}c^{k+1}$
$= b^{2^k+2^k}c^{k+1}$
$= b^{2^{k+1}}c^{k+1}$

So we have that $c^nb=b^{2^n}c^n$ . This implies $c^{16}ba^2 = b^{2^{16}}c^{16}a^2$ . Since we showed earlier that $a^4cb = b^2c^2a$ , we have
$b^2c^2a = a^4cb = b^{2^{16} }c^{16}a^2.$
So
$a = (b^{2^{16} }c^{16})^{-1}b^2c^2 = c^{-16} b^{2-2^{16}} c^2 .$
$c^2= a c a^{-1}$
$= c^{-16} b^{2-2^{16}} c^2 \cdot c \cdot (c^{-16} b^{2-2^{16}} c^2 )^{-1}$
$= c^{-16} b^{2-2^{16}} c^2 \cdot c \cdot (c^{-2} b^{2^{16}-2} c^{16} )$
$= c^2= c^{-16} b^{2-2^{16}} \cdot c \cdot b^{2^{16}-2} c^{16}$
$= b^{2-2^{16}} \cdot c \cdot b^{2^{16}-2}$
$= b^{2(1-2^{15})} \cdot c \cdot b^{2(2^{15}-1)}$

Since $cb=b^2c$ , $cb^n=b^{2n}c$ . This can be shown by induction.
We have the base step. Induction step, assume true for $k$ : $cb^{k} = b^{2k}c$ . Then
$cb^{k+1} = cb \cdot b^{k} = b^2c \cdot b^{k} = b^2 (cb^k) = b^2 b^{2k}c = b^{2(k+1)}c.$
So
$c^2= b^{2(1-2^{15})} \cdot c \cdot b^{2(2^{15}-1)} = = b^{2(1-2^{15})} \cdot b^{4(2^{15}-1)}c$
implies $c = b^{2(2^{15}-1)}$ . So $b$ and $c$ commute since $c$ is generated by $b$ .
$b^2=cbc^{-1}=cc^{-1}b$ implies $b$ is the identity because $b^2=b$ .
Since $b$ is the identity, then so is $c$ and so is $a$ .
$G$ is the trivial group with a very complicated group presentation.

Written by bs8277

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