Problem 5.2.12 (Central product $Zn * Z_m$)

Question

Let $n$ and $m$ be positive integers with $d = (n, m)$. Let $Z_n = \langle x \rangle$ and $Z_m = \langle y \rangle$. Let $A$ be the central product of $\langle x \rangle$ and $\langle y \rangle$ with an element of order $d$ identified, which has presentation $\langle x, y \mid x^n = y^m = 1,\ xy = yx,\ x^{\frac{n}{d}} = y^{\frac{m}{d}} \rangle$. Describe $A$ as a direct product of two cyclic groups.

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $d = \mathrm{g.c.d.}(n, m)$. The cyclic group $Z_n = \langle x angle$ contains the element  $x^{n/d}$ of order $d$, and similarly $Z_m = \langle y angle$ contains the element  $y^{m/d}$ of order $d$. Thus $Z_n = Z(Z_n)$ contains a subgroup $Z_1 = \langle x^{n/d} angle$, and $Z_m = Z(Z_m)$ contains a subgroup $Z_2 = \langle x^{m/d} angle$ such that $Z_2 \simeq Z_1$, where the isomorphism $\varphi$ is characterized  by $\varphi(x^{n/d}) = y^{m/d}$, so that $\varphi( x^{k n/d}) = y^{km/d}$ for all integers $k$. 

By definition, the central product of $Z_n$ and $Z_m$ is a quotient
$$Z_n * Z_m = (Z_n 	imes Z_m)/ Z \qquad 	ext{where} \qquad Z = \{(x^{k n/d}, y^{-km/d}) \mid k \in [\![0,d[\![\ \}.$$
As in Exercise 5.1.13, we can prove that $Z_n * Z_m$ has presentation
$$Z_n * Z_m \simeq \langle x, y \mid x^n = y^m = 1,\ xy = yx,\ x^{\frac{n}{d}} = y^{\frac{m}{d}} angle.$$
Note that
$$|Z_n * Z_m | = \frac{|Z_n|\cdot |Z_m|}{|Z|} = \frac{nm}{d}.$$

Furthermore, since $d = \mathrm{g.c.d.}(n,m)$, there are integers $u,v$ such that
$$u \frac{n}{d} - v  \frac{m}{d} = 1.$$

If $x,y$ denote the generators of $G$, where  
$$G = \langle x, y \mid x^n = y^m = 1,\ xy = yx,\ x^{\frac{n}{d}} = y^{\frac{m}{d}} angle,$$
 then
\begin{align*}
x &= x^{u \frac{n}{d} - v  \frac{m}{d}} \
&= \left(x^{\frac{n}{d}} ight)^u x^{-v \frac{m}{d}}\
&=\left(y^{\frac{m}{d}} ight)^u x^{-v \frac{m}{d}}\
&=(x^{-v} y^u)^{\frac{m}{d}},\
&\
y &= y^{u \frac{n}{d} - v  \frac{m}{d}} \
&=y^{u \frac{n}{d}} \left(y^{\frac{m}{d}}ight)^{-v}\
&=y^{u \frac{n}{d}} \left(x^{\frac{n}{d}}ight)^{-v}\
&=(x^{-v} y^u)^{\frac{n}{d}}.
\end{align*}
This proves that
$$G = \langle x, y angle = \langle x^{-v} y^u angle,$$
So $G$ is cyclic, of order $\frac{nm}{d}$. In conclusion
$$Z_n * Z_m \simeq Z_{\frac{nm}{d}}.$$

\end{proof}
Note: the answer $Z_n 	imes Z_{m/d}$ is false, since this group is isomorphic to $Z_{\frac{nm}{d}}$ only if $\mathrm{g.c.d}(n, m/d) = 1$.

Example: for $n = 6, m = 9$,
\begin{align*}
G &= \langle x, y \mid x^6 = y^9 = 1, \ xy = yx, \ x^2 = y^3 angle\
&=\{1,x,x^2,x^3,x^4,x^5,y, xy, x^2 y ,x^3y, x^4y, x^5y, y^2, xy^2, x^2y^2, x^3 y^2, x^4 y^2, x^5 y^2\}\
&\simeq Z_6 * Z_9
\end{align*}
is generated by $xy^2$, so is isomorphic to $Z_{18}$, but is not isomorphic to $Z_6 	imes Z_3$.

Problem 5.2.12 (Central product $Zn * Z_m$)

Answers

Comments

Problem 5.2.12 (Central product $Zn * Z_m$)

Answers

Comments

Add answer