Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose I have two presentations for groups:

$\langle x,y|x^{7} = y^{3} = 1, yx = x^2y\rangle$ and $\langle x,y|x^{7} = y^{3} = 1, yx = x^4y\rangle$

What is the standard approach to deciding whether the presentations are isomorphic?

I'm working through an application of Sylow Theory which classifies groups of order $21$.

In the text it says that these two presentations above are isomorphic, but I cannot see how to prove it or even suspect it.

share|cite|improve this question
up vote 3 down vote accepted

Well, you'd want to find a set of generators of the first group that satisfied the relations of the second group. If we rewrite $yx=x^2y$ as $yxy^{-1}=x^2$ (which turns this somewhat abstract equality into something a bit more concrete), we see immediately that $y^2xy^{-2}=x^4$ and indeed since $y^2$ is of order 3, $x,y^2$ are the generators you're looking for.

share|cite|improve this answer
I think you mean $yx^2y^{-1}=x^4$, and that you would mean $y^2$ is of order $3$ but that instead you'll change that based on the other typo to $x^2$ is of order $7$. – Kevin Carlson Oct 12 '12 at 23:19
No, that's not what I meant. If you take $x^2$ and $y$ as generators, you get the same relations back. The stated equality is correct (although maybe I shouldn't have used the word "immediately"); the idea is simply that if conjugating by $y$ corresponds to squaring, then conjugating twice should correspond to taking a fourth power. – anonymous Oct 12 '12 at 23:31
Yep, you're right on all material counts-my deleted answer looks to have constructed the inverse of what I meant it to. Don't you still want to say $y^2$ is of order $3$ to guarantee you can use it as a generator? – Kevin Carlson Oct 12 '12 at 23:46
That was the intended implication but the argument is certainly clearer (and equally short) if I say $y^2$ is of order 3, so I have edited my answer accordingly. – anonymous Oct 12 '12 at 23:50
Thanks for the great answer. – roo Oct 13 '12 at 0:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.