Finding a normal and not normal subgroup of $S_3$
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?
$\endgroup$ 35 Answers
$\begingroup$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?
$\endgroup$ 1 $\begingroup$$S_3 =\{1 , (12),(13),(23),(123),(132)\}$,then $A_3 =\{1 , (123),(132)\}$ is normal subgroup of $S_3$ and $\{1 , (12)\} ,\{1,(13)\} ,\{1,(23)\} $ are non normal subgroups of $S_3.$
$\endgroup$ 1 $\begingroup$$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.
$\endgroup$ $\begingroup$$H_1=\{e,(1 2)\}$ is not a normal subgroup of $S_3$. However, $H_2=\{e,(1 2 3),(1 3 2)\}$ is a normal subgroup of $S_3$.
$\endgroup$ 4 $\begingroup$Normal subgroup of $S_3$ are $\{e\}$, $\{e, (123), (132)\}$, and $S_3$ itself. All subgroups with two elements are not normal.
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