# Is the given group $H$ is a normal subgroup of $S_4$?

I need to show that the group $H=\{(1),(12)(34),(13)(24),(14)(23)\}$ is a normal subgroup of $S_4$.

I know that I have to show $gH=Hg$ for every arbitrary element $g$ of the group $S_4$.

I have chosen $(123)$ as $g$.

For the left coset I obtain: $(123)(12)(34)=(134)$.

For the right coset I obtain: $(12)(34)(123)=(243)$.

I have found different results; there is a problem, but where?

Where am I wrong? Thanks for your help.

A two permutation in $S_n$ are conjugate to each other if and only if they have the same cycle structure.Now note that in $S_4$ number of elements of the form (1 2)(3 4),(1 3)(2 4),(1 4)(2 3) is 3 and these are all.so Any conjugate of these elements will be among themselves. Now your H contains all the conjugates.So H is normal in $S_4$. If you still have any doubt feel free to ask.
You have to verify $gHg^-1=H$ for all $g \in S_4$. The good news is that $S_4$ is generated by $(1,2)$ and $(1,2,3,4)$ so you only have to verify this for these $g$.
To prove $H$ is normal in $G$, we only need to check that $gH=Hg$, for all $g \in G$. We do not need to satisfy the stronger condition that $gh=hg$, for all $g \in G, h \in H$.
Equivalently, we need to prove that $g^{-1}Hg=H$, for all $g \in G$. In the special case $G$ is the symmetric group, the permutation $g^{-1}hg$ has the same cycle structure as $h$. Thus, for a subgroup $H$ of $S_n$ to be a normal subgroup, it is necessary and sufficient that $H$ satisfy the following condition: if $H$ contains a permutation $h$, then $H$ contains all permutations of $S_n$ that have the same cycle structure as $h$. There are exactly three permutations in $S_4$ that are a product of two disjoint transpositions and $H$ contains all of them. Thus the subgroup $H$ is a union of conjugacy classes of $G$, hence is normal.