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If I know that a group of finite order has two elements $a$ and $b$ s.t. their orders are $6$ and $10$, respectively. What statements can be made regarding the order of the group?

I know by Lagrange's that the elements should divide the order of the group, so I've taken the $\operatorname{lcm}$. I think the order of our group should be a multiple of $30$. But I'm thinking there's more I can say.

Alexander Gruber
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    Is your hypothesis that $a^6=b^{10}=e$, or that the order of $a$ is 6, and the order of $b$ is $10$? – Martin Argerami Jun 07 '13 at 02:41
  • On the assumption that the orders of the elements are $6$ and $10$ can you see how to construct a group of order $30n$ for $n\in \mathbb N$ – Mark Bennet Jun 07 '13 at 02:45
  • @MartinArgerami Aren't they saying the same thing? Sorry, I was trying to be more formal about it. Thanks for pointing it out – TheRealFakeNews Jun 07 '13 at 02:46
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    @AlanH For $a$ to have order $6$, we must have that $6$ is the smallest number $n>0$ for which $a^n=1$. Consider the identity, for example - surely $a=1$ satisfies $a^n=1$ no matter which $n$ we choose! – Alexander Gruber Jun 07 '13 at 02:50

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You can't say anything about the order of the group except that it must be divisible by $6$ and $10$. In fact, nothing can be said about the order of $ab$, either! (see the commentary here.)

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Alexander Gruber
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  • could we say it has an element of order 2, 3, and 5? $x^6 = (x^2)^3 = (x^3)^2$, and similarly for $x^{10}$. I know that doesn't say much about $G$, but I'm just trying to figure out what more can be said. – TheRealFakeNews Jun 07 '13 at 03:03
  • @AlanH If you know that $x$ has order $6$, then $x^2$ surely has order $3$ and $x^3$ surely has order $2$. Can you can generalize that to $x^{n}$? Furthermore, can you generalize that to prove that a group in which every nontrivial element has the same order must be a $p$-group (a group of prime power order)? – Alexander Gruber Jun 07 '13 at 04:31
  • To generalize, do I just take all divisors of $n$? If every nontrivial element has the same order, then it must be a $p$-group because the only divisors of $p$ are $1$ and $p$ itself (this means the group $G$ in my problem is certainly not a $p$-group). Is this correct? – TheRealFakeNews Jun 08 '13 at 02:14