How do I prove that the subset of unfixed transpositions from $A_4$ is normal to $A_4$? Let $K = \{\epsilon,(12)(34),(13)(24),(14)(23)\}$ 
How can I show that $gK=Kg$ for all $g\in A_4$?
Without using the kernel is isomorphic to the the sign function, how would 
I prove that $K \trianglelefteq A_4?$ 
I understand that $K$ is the set of all transpositions in which no element is fixed. While playing around with the left and right cosets I observed that the composition of $g_1 \in A_4\setminus K$ with $K$ of the form $g_1K$ and $Kg_1$ resulted in the set $A_4\setminus K$. Similarly since $K\leq A_4$ and $K$ is abelian for any $g_1\in K$ then $g_1K=Kg_1=K$. It follows that $K\cup ( A_4 \setminus K) = A_4$ so for $g_1 \in A_4$ $g_1K=Kg_1=A_4$ so in fact $K\trianglelefteq A_4$.
Note:
$A_4\setminus K = \{(123),(132),(124),(142),(134),(143),(234),(243)\}$
I have observed as well that the composition of two 2-cycles with one 3-cycle and vise versa results in a 3-cycle.
 A: $gK=Kg$ is equivalent to $gKg^{-1}=K$. This means that $gag^{-1}\in K$
for all $a\in K$. But $gag^{-1}$ is a conjugate of $a$, and conjugates
have the same cycle structure. So $gag^{-1}$ is either the identity, or
has two cycles of length two. Either way, $gag^{-1}\in K$.
A: First of all, you want to describe $K$ by saying that $K\setminus\{e\}$ is the set of all involutions with no fixed point.  (Each of them happens to be a product of two commuting transpositions.  However, the transpositions themselves are not even in $A_4$.  A transposition is a permutation that moves exactly two points.)  The easiest way to see that $K$ is normal in $A_4$ is to see that it is a union of conjugacy classes.  But $e$ is its own conjugacy class, and all conjugates of permutations with a $2^22^2$ cycle structure have the same cycle structure, so that rest of $K$ is also a union of conjugacy classes (in fact $K\setminus\{e\}$ is a single conjugacy class as you can easily see but you don't need that).  But a subgroup is normal if and only if it is a union of conjugacy classes.  So $K$ is normal in $A_4$.  QED
