0
$\begingroup$

I think it's false and here's is my work

let $$a = \left(\begin{array}{rr}0& 1\\ -1& 0\end{array}\right)$$ and $$b = \left(\begin{array}{rr}\sqrt {-1}& 0\\ 0&-\sqrt {-1}\end{array}\right).$$ then, former subgroup is

${\{\epsilon, -\epsilon, a, -a, ab, -ab, a^2, a^3, b, b^2, b^3, -b\}}$ so the order is 12.

Since isomorphic group has same order, the latter one should have order 12, that is $A_4$

This's all what I can do.. What should I do to prove it?

$\endgroup$
5
  • $\begingroup$ I think the generated subgroup is of order eight only. You can then use the fact that the subgroups of order $8$ in $S_4$ (all isomorphic to the dihedral group) have several elements of order two, but this group has only one, namely $-I_2$. The elements of this subgroup other than $\pm I_2$ all have order four. $\endgroup$ Apr 15, 2017 at 5:46
  • $\begingroup$ For example $b^3=-b$ and $b^2=-\epsilon=a^2$. $\endgroup$ Apr 15, 2017 at 5:58
  • $\begingroup$ @JyrkiLahtonen Oh, I miscalculated that!... :P Thanks for your answer :) $\endgroup$
    – Orange
    Apr 15, 2017 at 10:44
  • $\begingroup$ @JyrkiLahtonen I'm curious if their orders are same, should I calculated them all one by one till I found whether it is isomorphic? $\endgroup$
    – Orange
    Apr 15, 2017 at 10:45
  • $\begingroup$ you have a mistake in calculating the order - $a^2=b^2=-\epsilon$, $a^3=-a$, $b^3=-b$ so the order is $8$. I believe this is isomorphic to the quarternion group. $\endgroup$ Apr 16, 2017 at 9:12

1 Answer 1

1
$\begingroup$

You're correct that it is false. I'll give a sketch proof, the details shouldn't be too hard to fill.

Let $G$ be the group genereated by $a,b$. You can check (there's only 8 elements, so you can check this by hand) that any non-trivial subgroup of $G$ contains $-I_2$.

Suppose $G$ is isomorphic to a subgroup of $S_4$ - this is equivalent to saying $G$ acts faithfully on $\Omega=\{1,2,3,4\}$. Consider the stabiliser $G_x$ for some $x\in\Omega$. By the orbit-stabiliser theorem, $|G_x|=|G|/|x^G|\ge 2$ (where $x^G$ is the orbit of $G$ containing $x$).

In particular $G_x$ is non-trivial so contains $-I_2$ for each $x\in\Omega$. That is $I_2$ fixes every $x\in\Omega$ contrary to the assumption that $G$ acts faithfully.

$\endgroup$
1
  • $\begingroup$ It's little hard to understand but Thanks Robert :) I will try to understand it. $\endgroup$
    – Orange
    Apr 17, 2017 at 2:03

You must log in to answer this question.

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