Computation of Amalgamated Product $\mathbb{Z}_4 \ast_{\mathbb{Z}_2} \mathbb{Z}_6.$

Question

I'm trying to compute a amalgamated product $\mathbb{Z}_4 \ast_{\mathbb{Z}_2} \mathbb{Z}_6$.

Let $\mathbb{Z}_4= \langle a\mid a^4 =1\rangle$ and $\mathbb{Z}_6 = \langle b\mid b^6 =1\rangle $, be a representations of additive groups $\mathbb{Z}_6$ and $\mathbb{Z}_4$.

The free group $\mathbb{Z}_4 \ast \mathbb{Z}_6$ can be write in the following form

$$ \mathbb{Z}_4 \ast \mathbb{Z}_6 = \langle a,b\mid a^4=1,b^6=1\rangle $$

Let, $\varphi_1 : \mathbb{Z}_2 \to \mathbb{Z}_4$, $a \mapsto a^2 $ and $\varphi_2 : \mathbb{Z}_2 \to \mathbb{Z}_6$, $b \mapsto b^3$. Observe that, $\varphi_1$ take a element of order 2 and carry to a element of order 2, then $\varphi_1$ is injective. By the same way, we have that $\varphi_2$ is injective.

Then $$ \mathbb{Z}_4 \ast_{\mathbb{Z}_2} \mathbb{Z}_6 =\langle a,b\mid a^4=1,b^6=1, a^2 = b^3 \rangle = \langle b\mid b^6 =1 \rangle. \tag{$\star$} $$

Are my calculations correct?

Edit

I saw a discussion on Shaun post Presentations of Amalgamated Free Products of Two Groups., and I think that I understand my confusion on the notation and about my doubt on the comment below.

In fact, $\left( \star \right)$ should be written as

$$ \mathbb{Z}_4 \ast_{\mathbb{Z}_2} \mathbb{Z}_6 =\langle a,b\mid a^4,b^6, a^2 b^{-3} \rangle = \langle b\mid b^6 \rangle $$

where, $a^2b^{-3}$ is a word.

Answer 1

First of all, even your definition of the amalgam $G={\mathbb Z}_4\star_{{\mathbb Z}_2} {\mathbb Z}_6$ is wrong. You have to give a name, say, $c$ to the generator of the amalgamated subgroup ${\mathbb Z}_2$. Then you can say that the monomorphisms ${\mathbb Z}_2\to {\mathbb Z}_4$ and ${\mathbb Z}_2\to {\mathbb Z}_6$ are given, respectively, by $$ c\mapsto a^2, \ c\mapsto b^3. $$ Then you correctly identified the presentation for the corresponding amalgam as $$ \langle a,b\mid a^4,\ a^2b^{-3},\ b^6\rangle $$ (no need to write $=1$). What you wrote next, however, is completely unjustified: $$ \langle a,b\mid a^4,b^6, a^2 b^{-3} \rangle = \langle b\mid b^6 \rangle. $$ Your attempted justification in the comment goes absolutely nowhere; you write: $$ ^2=^3 \Rightarrow (^2)(^2)=^6=1$$ In fact, the two implied equalities do not follow from $a^2=b^3$, but do follow from $a^4=b^6=1$. What's worse, none of this explains how did you get rid of the generator $a$! Actually, one can easily see that this simply cannot be done as one can construct an epimorphism from $G$ (with the presentation $\langle a,b\mid a^4,b^6, a^2 b^{-3} \rangle$) to the free product $$ H={\mathbb Z}_2\star {\mathbb Z}_3, $$ by sending $a$ to the generator of the first free factor and $b$ to a generator of the second free factor. Since $H$ is non-abelian, so is $G$.

Answer 2

You're wrong.

According to GAP:

gap> F:=FreeGroup(2);
<free group on the generators [ f1, f2 ]>
gap> rels:=[ (F.1)^4, (F.2)^6, (F.1)^2*(F.2)^(-3) ];
[ f1^4, f2^6, f1^2*f2^-3 ]
gap> G:=F/rels;
<fp group on the generators [ f1, f2 ]>
gap> IsCyclic(G);
false

That is, the group cannot be cyclic and so is not isomorphic to $\Bbb Z_6$.

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Inspired by @MoisheKohan's answer here:

Kill $a^2$ and $b^3$ in

$$\begin{align} G&:=\mathbb{Z}_4 \ast_{\mathbb{Z}_2} \mathbb{Z}_6 \\ &\cong\langle a,b\mid a^4,b^6, a^2b^{-3} \rangle. \end{align}$$

Then, homomorphically, $G$ maps onto

$$\begin{align} H&\cong \langle a,b\mid a^2,a^4, b^3, b^6, a^2b^{-3}\rangle\\ &\cong\langle a,b\mid a^2, 1^2, b^3, 1^2, 1(1)^{-1}\rangle\\ &\cong\langle a,b\mid a^2, b^3\rangle\\ &\cong\Bbb Z_2\ast\Bbb Z_3, \end{align}$$

which is nonabelian. Hence $G$ is nonabelian, so is not cyclic. Hence $G\not\cong\Bbb Z_6$.