# How to show the Symmetric Group $S_4$ has no elements of order $6$.

I have just shown that $\rho$, where $\rho$ is a product of disjoint cycles of lengths $m_i$ in $S_n$, has order $\operatorname{lcm}(m_1,m_2,\ldots,m_k)$.

Now I have to show that there are no elements of order $6$ in $S_4$. I know that $S_4$ has order $24$, but I can't really get anywhere from this. Could someone give me a hint on how to do this. Thanks.

You need $\operatorname{lcm}(m_1,\ldots,m_k)=6$ and $m_1+\cdots+m_k=4$. That second equality admits only a few solutions: \begin{align} & 4 \\ & 3+1 \\ & 2+2 \\ & 2+1+1 \\ & 1+1+1+1 \end{align}

So find the $\operatorname{lcm}$ in each of those cases.

Hint: write a candidate as the product of disjoint cycles and use your first result. Since there are only four elements, the possible cycle lengths are quite limited; e.g. four cycles of length one, one cycle of length 3 and one cycle of length one, etc.

for $\alpha \in S_4$ to be of order $4$:

you need to write $\alpha$ as product of disjoint cycles whose lenghts has l.c.m $6$.

Max. No. of elements in any element of $S_4$ would be $4$.

You can not write $4$ as $a+b+c+d$ such that l.c.m of $a,b,c,d$ is $6$ (which you have to prove)

We know that any permutation in $S_n$ is a product of disjoint cycles. And order of an $r$ cycle is $r$. The order of any permutation is lcm of the order of all $r$-cycles in it. We shall require these things.

First of all, if we need to search element of order 6 in $S_4$ then it is (by the above logics) either a 6 cycle or disjoint product of 2-cylce and 3-cycle. Why? because in the first case then order is directly becoming 6 and for the second case, the order is lcm$(2, 3)=6$.

But since $S_4$ is nothing but group of all permutations on 4 distinct symbols, so using 4 symbols we can't create 6-cycle. First case is thus cancel out. For the second case, if we choose any 3-cycle then we are left with 1 symbols using which we can't create any 2-cycle.

Hence no element of order 6 is there in $S_4$.