Does the permutation group S_8 contain elements of order 14?

In summary, the permutation group $S_8$ does not contain elements of order 14, as the only possible disjoint cycle decompositions in $S_8$ with a length of 14 are $(14)$ and $(2,7)$, both of which are not possible due to the limitations of the group. This is because 14 has the prime factorization of $2\cdot 7$, and for an element to have order 14, it must either be a 14-cycle or a combination of a 2-cycle and a disjoint 7-cycle. However, both of these options are not possible in $S_8$ as they would require a total of 9 elements, which exceeds the order of the group
  • #1
i_a_n
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Does the permutation group $S_8$ contain elements of order $14$?My answer: If $\sigma =\alpha \beta$
where $\alpha$ and $\beta$ are disjoint cycles, then
$|\sigma|=lcm(|\alpha|, |\beta|)$ .
Therefore the only possible disjoint cycle decompositions for a permutation $\sigma \in S_8$ with $|\sigma| =14$ is $(7,2)$. Since $7+2\neq 8$ so there is no element of order 14 in $S_8$.

Is my answer right?
 
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  • #2
Re: Does the permutation group $S_8$ contain elements of order 14 and the number of those element?

ianchenmu said:
Does the permutation group $S_8$ contain elements of order $14$?My answer: If $\sigma =\alpha \beta$
where $\alpha$ and $\beta$ are disjoint cycles, then
$|\sigma|=lcm(|\alpha|, |\beta|)$ .
Therefore the only possible disjoint cycle decompositions for a permutation $\sigma \in S_8$ with $|\sigma| =14$ is $(7,2)$. Since $7+2\neq 8$ so there is no element of order 14 in $S_8$.

Is my answer right?I feel I've only considered the possibility that
[FONT=MathJax_Math]σ[/FONT] is product of two disjoint cycles but I don't know how to count the number of elements of order 14 in a single cycle. So what's the right answer?

Your answer is right... with a couple of tweaks.

Since 14 has the prime factorization $2 \cdot 7$ there are only two permutation types possible for an element of order 14.
Either it must be a 14-cycle, or it has to be the combination of a 2-cycle and a disjoint 7-cycle.

It is not so much that $7+2\neq 8$, but it is not possible because $7+2 > 8$.
And of course 14 is also greater than 8.
To clarify, $S_8$ does have an element of order 12: a 3-cycle combined with a disjoint 4-cycle.
This works because $3+4 \le 8$ and because 3 and 4 are relatively prime.
 

FAQ: Does the permutation group S_8 contain elements of order 14?

What is the permutation group S_8?

The permutation group S_8, also known as the symmetric group of degree 8, is a group that consists of all possible permutations of 8 elements, where a permutation is a rearrangement of the elements in a specific order.

What does it mean for an element to have an order of 14 in a permutation group?

The order of an element in a permutation group refers to the smallest number of times that the element needs to be applied to itself to return to its original position. In other words, it is the number of elements in the subgroup generated by that element.

Can elements of order 14 exist in the permutation group S_8?

Yes, elements of order 14 can exist in the permutation group S_8. In fact, S_8 contains elements of all possible orders, including 14. However, not all elements in S_8 will have an order of 14.

How can we determine if an element in S_8 has an order of 14?

To determine the order of an element in S_8, we can use the following formula: order = lcm(cycle lengths), where lcm stands for the least common multiple. This means that we need to find the cycle lengths of the element's permutation and then find the least common multiple of those lengths. If the result is 14, then the element has an order of 14.

Can we find examples of elements of order 14 in S_8?

Yes, we can find examples of elements of order 14 in S_8. For example, the permutation (1 2 3 4 5 6 7 8)(9 10 11 12 13 14) has an order of 14 in S_8. This means that when we apply this permutation to itself 14 times, we will get back to the original arrangement.

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