The question is:
Consider $S_4$, the group of all permutations on $A = \{1,2,3,4\}$, and let $G$ be the subgroup consisting of all even permutations in $S_4$.
How would I show that $G$ has no subgroup $H$ with 6 elements? I'm not getting what it means by elements... wouldn't an element of $G$ be an even permutation?

  • $\begingroup$ Here is a list of all the subgroups of $G$ (AKA $A_4$). $\endgroup$ – Kenny Lau May 13 '17 at 10:12

Idea for you to think about:

Suppose $\;H\le A_4\;,\;\;|H|=6\;$ , so there are two possibilities:

=== $\;H\cong S_3\;$ . Check now the elements of order two and three in $\;A_4\;$ and rule this out (further, possibly helpful, hint: the set of all elements in $\;A_4\;$ of order two, together with the identity element, is a normal subgroup even of $\;S_4\;$ and thus, of course, also of $\;A_4\;$ ).


=== $\;H\cong C_6\;$ . This is even easier: check there is no element of order six in $\;S_4\;$ , let alone in $\;A_4\;$ .


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