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Let $G=<a,b | a^2b=ba^2, b^2a=ab^2, a^2=b^4>$. Consider a homomorphism $\phi$ from $G$ to $<c| >$ where $\phi(a)=c^2$ and $\phi(b)=c$. Can someone help me decide what group the kernel of this map is? Thanks!

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The group can be written as a nice semidirect product, see here:… – user641 Feb 10 '11 at 1:30
In particular the kernel is the infinite dihedral group, generated by $ab^{-2}$ and $bab^{-3}$. – user641 Feb 10 '11 at 1:32

This may help to understand this group:

Every element can be written as $a^{i_1} b^{j_1} a^{i_2} b^{j_2}\cdots$.

Now since $a^2$ and $b^2$ both commute with both $a$ and $b$ we can collect even powers on one side, for instance $a^7b^3ab = abab\cdot a^6b^2$. Now substituting $a^2$ by $b^4$ on the right, on obtains an expression of the form $ b^\epsilon (ab)^n a^{\epsilon^\prime} b^{2\ell}$, where $\epsilon,\epsilon^\prime \in \{0,1\}$ and $n \in \mathbb N$ and $\ell \in \mathbb Z$.

Now it should be easy to see which of these elements are in the kernel of the map.

[edit] I've given it some more thought, this might help too. I haven't double checked the details yet though so there may be a mistake somewhere.

The kernel consists of elements of this form $$ (ab)^{2x} b^{-6x} , (ab)^{2x} a b^{-6x-2}, (ba)^{2x} b^{-6x}, (ba)^{2x+1}b^{-6x-3}$$ Note for instance that $(ab)^{2x} a b^{-6x-2} = (ab)^{2x+1} b^{-6x-3}$ so one can collect these into $$ (ab)^n b^{-3n} , (ba)^n b^{-3n}$$ These are still related because $$ (ba)^{-1} = a^{-1} b^{-1} = ab b^{-2}a^{-2} = ab b^{-2}b^{-4} = ab b^{-6},$$ therefore with $m = -n$ we have that $$ (ba)^n b^{-3n} = (ba)^{-m} b^{3m} = (ab b^{-6})^m b^{3m} = (ab)^m b^{3m-6m} = (ab)^{-n} b^{3n}$$ therefore the kernel is the set of all $$ (ab)^n b^{-3n},$$ where $n \in \mathbb Z$. Denote $(ab)^n b^{-3n}$ by $g_n$, then we can attempt to determine the group law as function of $n$ and we find that

  • $g_u g_v = g_{u+v}$ if $u$ is even
  • $g_u g_v = g_{u-v}$ if $u$ is odd

This would imply the kernel is isomorphic to a group $(\mathbb Z,\circ)$ where $a\circ b = a + (-1)^a b$. Note that $2\mathbb Z$ is a normal subgroup and note that all odd numbers are involutions that act nontrivial on $2\mathbb Z$ by sending $a$ to $-a$.

I think that would make the exact structure $C_\infty \rtimes C_2$, where $C_\infty \cong (\mathbb Z,+)$ is the subgroup $2\mathbb Z$ and $C_2$ is (for instance) the subgroup ${0,1}\cong C_2$ acting nontrivially on $C_\infty$ by sending every element to its inverse.

[edit 2] And now I notice the comment underneath the question that points to a solution making this a bit pointless...

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Thanks for the answer. I agree that looking at the group like this will allow you to see what set the kernel is (as a set), but I can't tell what group the kernel is. For example, is the kernel $\mathbb{Z} \times \mathbb{Z}$ (or something)? – user6077 Feb 9 '11 at 21:01
I don't think the kernel is abelian, but infinite index subgroups are a pain for me to double check. If you change the range to <c|c^n>, then the (finite index) kernel is not even nilpotent, I don't think, but it is definitely not abelian. At any rate (e=1,n=1,e'=0,l=-2) and (e=0,n=0,e'=1,l=-1) seem to both be in the kernel and not commute, but the (e,n,e',l) normal form seems a little weird to me. – Jack Schmidt Feb 9 '11 at 23:03
Note that I continued my efforts a bit further. – Myself Feb 10 '11 at 21:08

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