0

How to prove the group is generated by $a, b$ is cyclic group, where $a^2=1, b^3=1 $ and $a,b \in \mathbb{C}$?

I guess the order of group $<a,b>$ is 6, and the elements are $0, ab, b^2, a, b, ab^2$. Is it right?

  • What about $ba$, for example? – lulu May 24 '18 at 13:05
  • If you add commutative (which you added when you said that $a,b\in\mathbb{C}$), then $(ab)^{-2}=b^{-2}=b$ and $(ab)^{3}=a^{3}=a$. Therefore, $ab$ generates the whole group. –  May 24 '18 at 13:09
  • @arugula $a$ and $b$ are cubic unit root in $\mathbb{C}$. –  May 24 '18 at 13:13
  • @arugula Why is it generated by $ab$ but not other forms such as $ab^2$? –  May 24 '18 at 13:21
  • 1
    @Tinzoe-Yui In principle it could be generated by other elements as well. That's is not a problem. For example, since it is generated by $ab$ it is also generated by $(ab)^{-1}=a^{-1}b^{-1}=ab^2$. –  May 24 '18 at 13:23
  • @arugula Is there a formula to calculate the order? –  May 24 '18 at 13:24
  • 1
    With the information given there are a few possibilities for the order. If $ab=1$, then $a=b=1$. So the order is $1$. If $b=1\neq a$, then $(ab)^2=b^{-1}=1$. Therefore, the order is $2$. If $a=1\neq b$, then $(ab)^2=b^{-1}\neq1$ and $(ab)^3=a=1$. Therefore, the order is $3$ in this case. Finally, if $a,b\neq1$, then $b^2\neq1$ too. Then $ab,(ab)^2=b^{-1},(ab)^3=a,(ab)^4=b^4=b\neq1, (ab)^5=ab\neq1$, and $(ab)^6=1$. Therefore, the order is $6$ in this case. –  May 24 '18 at 13:29

2 Answers2

1

This statement is not true without further constraints on the group. There is a group that is generated by two elements $a,b$, and that satisfies $a^2=1$ and $b^3=1$, but which is not a cyclic group. It is known as the free product $\mathbb{Z} / 2 \mathbb{Z} * Z / 3 \mathbb{Z}$ of the cyclic group $\mathbb{Z} / 2 \mathbb{Z}$ of order $2$ with the cyclic group $Z / 3 \mathbb{Z}$ of order $3$. This group is not even finite and is not abelian. In some sense this is the "free-est" or "most universal" group that is generated by two elements $a,b$ and that satisfies $a^2=1$ and $b^3=1$.

Lee Mosher
  • 135,265
1

If your group multiplication law is multiplication in $\mathbb C$ then the group is abelian (since multiplicaiton in $\mathbb C$ is commutative). This is all we need to show the group is cyclic.

If $a=1$, then the group is generated by $b$ only, so it is trivially cyclic; similarly if $b=1$, the group is generated by $a$ only so trivially cyclic. (Thanks to arugula for pointing out the need to consider this case.) Thus we assume $a \neq 1$ and $b \neq 1$.

In this case, the $6$ elements you list are indeed the only elements in the group, so to show the group is cyclic, you can demonstrate an element of order $6$. Clearly $0$ has order $1$, $a$ has order $2$, and $b$ and $b^2$ have order $3$. What about $ab$? $$(ab)^2 = a^2b^2 = b^2$$ and $$(ab)^3 = a^3b^3 = a$$ Neither of these is the identity, so the order of $ab$ is not $1$, $2$, or $3$. Since the order must divide the order of the group ($6$), it must be $6$. (Note that $ab^2$ is also an element of order $6$ -- similar reasoning would work.)

BallBoy
  • 14,740