0
$\begingroup$

I'm stumped by this question: Given $$ H = \langle (1,2)(3,4),(1,3)(2,4) \rangle \subset A_5 = G$$ deduce $\lvert N_G(H)\rvert$.

My attempt is to find a few elements of the normalizer and then use Lagrange's Theorem. So far I've shown that $(1,2,3) \in N_G(H)$ and so $\lvert N_G(H) \rvert$ must be a multiple of 12, and because $N_G(H) \leq A_5$ this multiple must divide 60. Hence $$\lvert N_G(H) \rvert = 12, 24, 36, 48, 60.$$ But at this point I'm stuck trying to find more elements of the normalizer, so I'm just guessing at permutations and computing to see if they're in the normalizer.

Is there a nicer way of doing this?

$\endgroup$
0

1 Answer 1

2
$\begingroup$

5 is the one and only element of $\{1,2,3,4,5\}$ not moved by any element of $H$. So, for any $g\in A_5$, $g(5)$ is the one and only element not moved by any element of $gHg^{-1}$. Therefore, if $g\in N_G(H)$, then $g(5)=5$. Now how many elements of $A_5$ satisfy $g(5)=5$?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .