# Show that all elements of order $42$ are conjugates in $S_{12}$.

I'm reading some lectures notes of my Abstract Algebra class and there is an example whose given solution I don't understand at all. I've been trying to solve it one my own without success, so I would appreciate if somebody can walk me through each step of this solution, or provide another (more thorough/detailed?) one.

I'm translating from russian.

PROBLEM:

Show that all elements of order $42$ are conjugates in $S_{12}$.

SOLUTION:

Let see which cycle type could have a permutation of the given order. The length of each cycle is a divisor of a permutation of order $42 = 2 \cdot 3 \cdot 7$.

It is necessary to find at least one cycle, such that its length is divisible by $2,\ 3$ and $7$ respectively. But notice that $2 + 3 + 7 = 12$. And also $2+3 < 2\cdot 3$, $2+7 < 2\cdot 7$, $3+7 < 3\cdot 7$, $2+ 3+7 < 2 \cdot 3\cdot 7$. So we have that for any choice of $2,\ 3$ and $7$ more than $12$ elements are needed in order to form the cycles.

This means that each permutation of order $42$ is formed by $3$ cycles of length $2,\ 3,\ 7$ respectively. That is, they all have the same cycle type and, therefore, conjugates.

• The checking is about excluding possibilities like two disjoint cycles of lengths $7$ and $6$. No room for those in $S_{12}$ as $7+6>12$. – Jyrki Lahtonen Mar 28 '17 at 20:32

## 1 Answer

The main thing is that

two permutations are conjugate in $S_n$ if and only if they have the same cycle decomposition type.

So the goal is to arrive that a permutation of order $42$ in $S_{12}$ necessarily has the cycle type $(2,3,7)$.

The possible cycle lengths are divisors of $42$, i.e. are among $$2,3,6,7,14,21,42$$ The last three are excluded because we are in $S_{12}$.

If we take too few cycles, (e.g. two disjoint cycles, one of order $2$ and $7$), then the order of their product is smaller than $42$ (in this example it's $2\cdot 7=14$).

Then, if instead of $(2,3,7)$, we would take disjoint cycles of length $6$ and of length $7$ to arrive at order $42$, then they again don't fit in $S_{12}$ as $6+7>12$.

Similarly handled all possibilities.