Take the 2-minute tour ×
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.

How can we show that the subgroup of $A_7$ generated by the permutations $x= (1234567)$ and $b=(26)(34)$ has order $168$?

share|improve this question
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. –  Julian Kuelshammer Dec 2 '12 at 19:24
For some basic information about writing math at this site see e.g. here, here, here and here. –  Julian Kuelshammer Dec 2 '12 at 19:25
related: math.stackexchange.com/q/1401/152 –  Grigory M Dec 3 '12 at 9:24

1 Answer 1

Assuming you want to avoid using a computer, you could observe that your two permutations are both automorphisms of the projective plane with 7 points and lines $\{1,2,6\}$, $\{1,3,4\}$, $\{1,5,7\}$, $\{2,3,7\}$, $\{2,4,5\}$, $\{3,5,6\}$, $\{4,6,7\}$. That would give you an upper bound of 168 on the order.

To get a lower bound, you could note for example that $b$ has conjugates $(37)(45)$ and $(25)(67)$ under powers of $a$, and then try and show that these two permutations together with $b$ generate a subgroup of order at least 24 stabilizing 1.

share|improve this answer

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.