4
$\begingroup$

I am trying to determine the group given by the presentation:

$\langle\ a, b, c\ \vert\ a^2=b^5,\ b^2=c^3,\ c^2=a^7\ \rangle$.

I've been trying to tackle this problem by starting with the first relation and, in essence, 'cycling' through the other two relations by substitution, in the hope that I produce some sort of simpler relation. So far, I've managed to deduce that

$a=b^{50}$ and $c^2=b^{65}$. These together with the original relation $c^2=a^7$ mean we can say: $b^{285}=1$.

From here, though, I am struggling to get any further. From what I've done so far, my intuition suggests to me that this is going to be some cyclic group, but I'm very unsure.

Any help would be appreciated!

  • 3
    We have $a^{8} = b^{20} = c^{30} = a^{105}$ which means that $a^{97} = 1$. This means that either $a = 1$ or $|a| = 97$. We can do the same starting at $b^8$ and $c^8$ giving $b^{97} = c^{97} = 1$ as well. Do you know that this group is abelian? – 2017-02-07
  • 0
    Hint: First show that this group is cyclic since 97 is a prime number. – 2017-02-07
  • 0
    @Arthur, thanks. I'm not assuming the group is abelian (But obviously it is). – 2017-02-07
  • 0
    You should check your calculation that $b^{285}=1$ which must be wrong because the group is isomorphic to $C_{97}$ with $b=a^{78}$ and $c=a^{52}$. – 2017-02-07
  • 0
    @DerekHolt Well, I haven't calculated the relation $b^{285}=1$ myself and I'm not sure if it is correct. But you say that it is isomorphic to $C_{97}$. Do you actually have a proof? Please provide one so we can have a closer look at the problem again. – 2017-02-08
  • 1
    As has been pointed out by Arthur and freakish, we have $a^{97}=b^{97}=c^{97}=1$. Since $2$ and $7$ are coprime to $97$, $a$ is a power of $a^2$ and of $a^7$, so each of $b$ and $c$ are powers of $a$. Hence $|G|=1$ or $97$. But the relations are satisfied with $a$ of order $97$, $b=a^{78}$ and $c=a^{52}$, so $|G|=97$. – 2017-02-08

1 Answers 1

2

As @Arthur mentioned $b^{8}=c^{12}=a^{42}=b^{105}$ so $b^{97}=1$. Now you know that $b^{285}=1$ thus

$$b^{285x + 97y}=1$$

for any $x,y\in\mathbb{Z}$. In particular, since $285$ and $97$ are coprime then there exist $x, y$ such that $285x+97y=1$. So $b=1$.

If $b=1$ then $a^2=1$ and $c^3=1$. Furthermore

$$1=c^3=cc^2=ca^7=ca$$

Thus $a = c^{-1}$. Therefore $c^2=1$ but $c^3=1$ implies $c=1$. On the other hand since $a=c^{-1}$ then $a^3=1$ and by the same argument $a=1$.

All in all your group is trivial.

  • 0
    I'm troubled. Surely the presentation is of the cyclic group $C_{97}$, and not just simply the trivial group. – 2017-02-07
  • 0
    @Joseph: Why do you think the group is $C_{97}?$ – 2017-02-07
  • 0
    Whoops just realised my mistake - just took $a^{97}=1$ to mean that $|a|=97$ – 2017-02-07
  • 0
    You have gone wrong somewhere, because the group is actually $C_{97}$. – 2017-02-07
  • 0
    @DerekHolt Please feel free to show me the mistake. I'm not saying that there is none but I don't see any. – 2017-02-08
  • 0
    @Joseph Yeah, that's not true. In general $a^n=1$ implies that $|a|$ divides $n$. – 2017-02-08
  • 1
    I can't show you the mistake, because the OP stated incorrectly that $b^{285}=1$. – 2017-02-08