MHB What Is the Structure of This Quotient Group?

[Deveno]

You are given a group as a quotient of the free group on two letters, a and b.

the kernel of the surjective homomorphism $F_2 \to G$ is generated by:

$\{a^7,b^6,a^4ba^{-1}b^{-1}\}$

a) prove $G$ is solvable by identifying the derived series:

$G' = [G,G] > G^{\prime \prime} = [G',G'] > \dots $

b) determine the isomorphism class of $G$

(hint: you have 6 to choose from)

 

[Opalg]

You are given a group as a quotient of the free group on two letters, a and b.

the kernel of the surjective homomorphism $F_2 \to G$ is generated by:

$\{a^7,b^6,a^4ba^{-1}b^{-1}\}$

a) prove $G$ is solvable by identifying the derived series:

$G' = [G,G] > G^{\prime \prime} = [G',G'] > \dots $ (I have removed a prime from [G,G'] to make it [G,G]. I hope that's right.)

b) determine the isomorphism class of $G$

(hint: you have 6 to choose from)

It is over 50 years since I took a course on group theory, so this has been a tough exercise for dormant brain cells.

From the relation $a^4ba^{-1}b^{-1} = e$ ($e$ being the identity element of $G$), it follows that $ba = a^4b$. So any word in $a$ and $b$ can be transformed to one of the form $a^kb^l$ by pushing all the $b$s to the right. Furthermore, each time you push a power $b$ past a power of $a$, the power of $b$ will not change, but the power of $a$ will be multiplied by 4. If you apply this process repeatedly to a commutator, the powers of $b$ will eventually all cancel and you will be left with a power of $a$. In particular, $aba^{-1}b^{-1} = a^4$, so $a^4$ is a commutator. Therefore $(a^4)^2 = a^8 = a \in [G,G]$, and it follows that $G' = \langle a\rangle$, the subgroup generated by $a$. Since that is abelian, the second derived group will just be $\{e\}.$

Next, $bab^{-1} = a^4$, which means that the subgroup $\langle b\rangle \cong \mathbb{Z}_6$ acts by inner automorphisms on the subgroup $\langle a\rangle \cong\mathbb{Z}_7$, and so $G$ is (isomorphic to) a semidirect product $\mathbb{Z}_6 \rtimes_{\alpha}\mathbb{Z}_7$, where $\alpha$ is the action given by $\alpha(a) = bab^{-1}.$ Use of the relation $ba = a^4b$ shows that $\alpha^3 = \text{id.}$, which makes it look as though the action of $\mathbb{Z}_6$ is effectively an action of $\mathbb{Z}_3$ together with a trivial action of $\mathbb{Z}_2.$ So I am guessing that $G$ is isomorphic to the direct product $G = H\times\mathbb{Z}_2$, where $H$ is the nonabelian group of order 21. But I don't have the technical knowhow to prove that properly.

 

[Deveno]

awesome. you're almost there!

let $H = \langle x,y: x^3 = y^7 = e,\ yx = xy^2\rangle$

it should be clear this gives a non-abelian group of order 21.

now you just have to pick a suitable choice for $a,b$ in $H \times \Bbb Z_2$ and show that these satisfy the defining relations of $G$.

OR:

eliminate the other 5 possibilities:

$G \cong \Bbb Z_{42}$ (but $G$ is non-abelian)
$G \cong D_{42}$ (but $G$ has an element of order 6)
$G \cong D_{14} \times \Bbb Z_3$ (has 7 elements of order 2)
$G \cong D_6 \times \Bbb Z_7$ (has 3 elements of order 2)
$G \cong \Bbb Z_2 \rtimes_{\phi} H$ (here H is the normal subgroup, for clarification).

the last one is tricky, to check the order of elements one needs to find an outer automorphism (which will be $\phi_1$) of $H$ of order 2 (clearly no inner automorphism will do).

perhaps:

$\phi_1(x) = x$
$\phi_1(y) = y^6$

might be a good choice (provided, of course, $y^6$ is not conjugate to $y$).

if one takes the latter approach, it might be fruitful to ask:

which elements of $G$ have order 2?

 

[Deveno]

so...can anyone confirm Opalg's guess?

 

[Deveno]

as hinted above, Opalg is indeed correct. let's prove this 2 ways, the easy way, and the hard way. the easy way first:

in $H \times \Bbb Z_2$, set:

$a = (y,0),\ b = (x,1)$

then $a^7 = (y^7,0) = (e,0)$ the identity, and

$b^2 = (x^2,0)$
$b^3 = (e,1)$
$b^6 = (e,1)^2 = (e^2,1+1) = (e,0)$

so $a$ is an element of order 7 and $b$ is an element of order 6.

now $ba = (x,0)*(y,0) = (xy,0)$

while $a^4b = (y^2,0)*(x,0) = (y^4x,0)$

recall that in $H$: $yx = xy^2$, so:

$y^4x = y^3(yx) = y^3(xy^2)= y^2(yx)y^2 = (y^2)(xy^4)$
$ = y(yx)y^4 = y(xy^2)y^4 = (yx)y^6 = (xy^2)y^6 = xy^8 = xy$

hence $a^4b = ba$.

*****************
now the hard way. we're going to count the number of elements of $G$ of order 2. one observation makes this somewhat easier:

$(a^jb^k)^2 = a^tb^{2k}$ for some integer t, since we don't alter the powers of b when we exchange $ba$ for $a^4b$, only the powers of a.

so if $(a^jb^k)^2 = e$ then $k = 3$, since $\langle a \rangle \cap \langle b \rangle = \{e\}$ (we can rule out $k = 0$ since $a$ is of order 7).

thus we have only six elements to check (we know $b^3$ is of order 2).

$(ab^3)^2 = ab^2(ba)b^3 = (ab^2a^4)b^4 = (ab)(ba)(a^3b^4)$
$= (ab)(a^4b)(a^3b^4) = (aba^4)(ba)(a^2b^4) = (aba^4)(a^4b)(a^2b^4)$
$= (aba)(ba)(ab^4) = (aba)(a^4b)(ab^4) = (aba^5)(ba)b^4 = (aba^5)(a^4b)b^4$
$= (aba^2)b^5 = a(ba)(ab^5) = a(a^4b)(ab^5) = a^5(ba)b^5 = a^5(a^4b)b^5$
$= a^2 \neq e$

$(a^2b^3)^2 = (a^2b^2)(ba)(ab^3) = (a^2b^2a^4)(ba)b^3 = (a^2b^2a)b^4$
$= (a^2b)(ba)b^4 = (a^2ba^4)b^5 = a^2(ba)(a^3b^5) = a^6(ba)(a^2b^5)$
$= a^3(ba)(ab^5) = bab^5 = a^4 \neq e$

$(a^3b^3)^2 = (a^3b^2)(ba)(a^2b^3) = (a^3b^2a^4)(ba)(ab^3)$
$= (a^3b^2a)(ba)b^3 = (a^3b^2a^5)b^4 = (a^3b)(ba)(a^4b^4)$
$= (a^3ba^4)(ba)(a^3b^4) = (a^3ba)(ba)(a^2b^4) = (a^3ba^5)(ba)(ab^4)$
$= (a^3ba^2)(ba)b^4 = (a^3ba^6)b^5 = a^3(ba)(a^5b^5) = (ba)(a^4b^5)$
$= a^4(ba)(a^3b^5) = a(ba)(a^2b^5) = a^5(ba)(ab^5) = a^2(ba)b^5 = a^6 \neq e$

$(a^4b^3)^2 = (a^4b^2)(ba)(a^3b^3) = (a^4b^2a^4)(ba)(a^2b^3) = (a^4b^2a)(ba)(ab^3)$
$= (a^4b^2a^5)(ba)b^3 = (a^4b^2a^2)b^4 = (a^4b)(ba)(ab^4) = (a^4ba^4)(ba)b^4$
$= (a^4ba)b^5 = a^4(ba)b^5 = a \neq e$

$(a^5b^3)^2 = (a^5b^2)(ba)(a^4b^3) = (a^5b^2a^4)(ba)(a^3b^3)$
$= (a^5b^2a)(ba)(a^2b^3) = (a^5b^2a^5)(ba)(ab^3) = (a^5b^2a^2)(ba)b^4$
$= (a^5b^2a^6)b^4 = (a^5b)(ba)(a^5b^4) = (a^5ba^4)(ba)(a^4b^4)$
$= (a^5ba)(ba)(a^3b^4) = (a^5ba^5)(ba)(a^2b^4) = (a^5ba^2)(ba)(ab^4)$
$= (a^5ba^6)(ba)b^4 = (a^5ba^3)b^5 = a^5(ba)(a^2b^5) = a^2(ba)(ab^5)$
$= a^6(ba)b^5 = a^3 \neq e$

finally (jeepers!):

$(a^6b^3)^2 = (a^6b^2)(ba)(a^5b^3) = (a^6b^2a^4)(ba)(a^4b^3)$
$= (a^6b^2a)(ba)(a^3b^3) = (a^6b^2a^5)(ba)(a^2b^3) = (a^6b^2a^2)(ba)(ab^3)$
$= (a^6b^2a^6)(ba)b^3 = (a^6b^2a^3)b^4 = (a^6b)(ba)(a^2b^4) = (a^6ba^4)(ba)(ab^4)$
$= (a^6ba)(ba)b^4 = (a^6ba^5)b^5 = a^6(ba)(a^4b^5) = a^3(ba)(a^3b^5)$
$= (ba)(a^2b^5) = a^4(ba)(ab^5) = a(ba)b^5 = a^5 \neq e$

this means that $b^3$ is the SOLE element of $G$ of order 2. this rules out 3 of the other 5 possibilities, and $G$ is non-abelian, so either:

$G \cong H \times \Bbb Z_2$, or:
$G \cong H \rtimes_\phi \Bbb Z_2$

of course to define the multiplication of the last group explicitly, we need to find an outer automorphism of $H$ of order 2.

it's clear that:

$\phi_1(x) = x$
$\phi_1(y) = y^6$

will be such an automorphism, provided $y^6$ is not some conjugate of $y$. so let's look at the conjugacy class of $y$.

note that:

$(x^k)(y^m)y(y^{-m})(x^{-k}) = (x^k)y(x^k)^{-1}$

so we need only conjugate $y$ by powers of $x$ to determine its conjugacy class. so:

$xyx^{-1} = xyx^2 = x(yx)x = x(xy^2)x = x^2y(yx) = x^2(yx)y^2= y^4$.
$x^2yx^{-2} = x(xyx^{-1})x^{-1} = xy^4x^{-1} = (xyx^{-1})^4 = y^{16} = y^2$.

so the conjugacy class of $y$ is $\{y,y^2,y^4\}$.

thus our automorphism is indeed outer. if we can establish that this semi-direct product group has more than one element of order 2, it cannot possibly be isomorphic to $G$.

clearly $(e,1)$ has order 2:

$(e,1)*(e,1) = (e\phi_1(e),1+1) = (e,0)$

and:

$(y,1)*(y,1) = (y\phi_1(y),1+1) = (y(y^6),0) = (e,0)$

and we are done (it turns out that ALL of the elements of the subgroup $\langle y \rangle \times \{0\}$ are of order 2).

of course, we might want to show that indeed $H \times \Bbb Z_2$ only has one element of order 2 as well, just as a sanity check:

suppose $(g,k)$ has order 2. this means:

$(g^2,k+k) = (0,e)$

since $H$ HAS no elements of order 2 (21 is odd), this forces $g = e$, so the only possibilities are:

$(e,1)$ and $(e,0)$

and only $(e,1)$ has order 2 (the other element is the identity).