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I'm being asked to find 2 subgroups of $S_3$, one of which is normal and one that isn't normal. I guess, to find the non normal subgroup is easier. I would do this by trial and error, but since the group is $S_3$ and easily visualizable, I guess that there should be a geometrical property that makes it easy to find a normal subgroup. Any of you guys know one? Or should I try this by brute force?

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  • $\begingroup$ The/a normal one is the alternating group $A_3$, I guess. I think this is the only normal subgroup other than {$e$}. Now you can look at a small mount of possibilities. $\endgroup$ – Gary. Jul 14 '15 at 23:32
  • $\begingroup$ For relatively small symmetric groups it's convenient to find an explicit form of their subgroups. For example, $S_{3}$ has 4 subgroups up to an isomorphism: trivail group, $S_{2}$, $A_{3}$, $S_{3}$ by itself. Then, it's straightforward to check, whether the group is normal or not. $\endgroup$ – hyperkahler Jul 14 '15 at 23:35
  • $\begingroup$ $S_3$ has only four proper nonidentity subgroups. One is normal and three are not. Trial and error really isn't bad. $\endgroup$ – Matt Samuel Jul 14 '15 at 23:36
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HINT: Keep in mind that $|S_3|=6$, so if you have any subgroup of order 3, it has index 2 and is therefore normal. Can you find a subgroup of order 3?

To find a nonnormal subgroup, you will need a different divisor of 6 that is not trivial, and the only choice there is $2$, so you are looking for a 2 element subgroup. Can you find that?

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$S_3$ is small enough that one can solve this by brute force, comparing left and right cosets of every subgroup. My suggestion: calculate all the left and right cosets of $H_1 = \{e, (1\ 2)\}$ and $H_2 = \{e, (1\ 2\ 3),(1\ 3\ 2)\}$, first.

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$S_3 =\{1 , (12),(13),(23),(123),(132)\}$,then $A_3 =\{1 , (123),(132)\}$ is normal subgroup of $S_3$ and $\{1 , (12)\} ,\{1,(13)\} ,\{1,(23)\} $ are non normal subgroups of $S_3.$

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