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The question is does $A_8$ have an element of order 26? My knowledge on alternating groups is very limited. I understand that the order of $A_8$ is $\frac{8!}{2}$ but that is essentially all. My one thought is maybe we use the theorem on cyclic subgroups, where the orders relate to $gcd$. Even that I am not sure how I would use though.

The teacher has not explained this and we are using the free book Abstract Algebra by Judson, which does not discuss orders of elements in alternating subgroups, only the order of the alternating subgroup. I am sorry if I have not provided enough information, I feel like with this question I am not sure where to begin.

Edit: We have not done Lagrange's yet

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    $\begingroup$ The order of an element must divide the order of the group $\endgroup$ – J. W. Tanner Feb 4 at 6:31
  • $\begingroup$ @J.W.Tanner Why are you answering in a comment? $\endgroup$ – Arthur Feb 4 at 6:32
  • $\begingroup$ @Arthur Okay, I posted an answer. $\endgroup$ – J. W. Tanner Feb 4 at 6:36
  • $\begingroup$ Just a note, we have not yet done Lagrange's yet, so I do not believe I can use this fact. $\endgroup$ – jesshn Feb 4 at 6:44
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The order of an element in a permutation group is the least common multiple of its cycle lengths. So, then, to get an element of order $26$, we would need a cycle of length $13$ or $26$ in there. In $A_8$, permuting eight elements, we can't do that.

In response to the request for clarification: If we take the $n$th power of a permutation expressed in terms of its cycle decomposition, that's the same as taking the $n$th power of each of its cycles. The $n$th power of a cycle of length $k$ is the identity if and only if $k$ divides $n$, and the $n$th power of the whole permutation is the identity if and only if the $n$th power of each of its cycles is the identity. The order of the permutation is the least positive $n$ for which this power is the identity, namely the least $n$ that's divisible by all of the $k$ - in other words, their least common multiple.

So, then, as an additional challenge exercise: what is the smallest $n$ for which the alternating group $A_n$ contains an element of order $26$?

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    $\begingroup$ Do you mind clarifying where this fact comes from: The order of an element in a permutation group is the least common multiple of its cycle lengths. I think I am struggling to understand what the order of an element in a permutation group is... $\endgroup$ – jesshn Feb 4 at 6:48
  • $\begingroup$ thank you for your explanations! $\endgroup$ – jesshn Feb 4 at 7:02
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The order of an element must divide the order of the group.

26 does not divide 8!/2, because 13 divides 26 but not 8!.

Therefore $A_8$ cannot contain an element of order 26.

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  • $\begingroup$ A nice simple answer, but unfortunately I can't use it because we have not proven that. Thank you though $\endgroup$ – jesshn Feb 4 at 7:16
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Its square would have order $13$, a prime. A permutation of prime order $p$ consists only of cycles of lengths $p$ and $1$ and must have a cycle of length $p$. If $p>n$ then there cannot be a permutation in $S_n$ with a cycle of length $p$. Of course, $13>8$.

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