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In my lectures, I've read that $A_5$ $($the alternating group of even length cycles in $S_5$$)$, has a subgroup of order $6$, and the example is: the group generated by $\langle (12) (34), (123)\rangle$.

I don't even understand what group we are generating? Wasn't a cyclic group only generated by one element? Why would this group be of order $6$?

Michael Hardy
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Nina
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    There are two elements there. One is $(1\ 2)(3\ 4)$ and the other is $(1\ 2\ 3)$. They don't generate a subgroup of order $6$. – Angina Seng May 11 '17 at 17:10
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    "Even length cycles" is not the right term. $A_5$ is the group of even permutations, which are those permutations that are the composition of an even number of transpositions. "Cycles", on the other hand" are a particular kind of permutation, and each cycle of even length is an odd permutation. – Michael Hardy May 11 '17 at 17:13
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    As Lord Shark the Unknown pointed out, that group is not of order six. Try the group generated by $(12)(45)$ and $(123)$ instead. The idea is that $(12)$ and $(123)$ generate a copy of $S_3$. When we use $(12)(45)$ instead of $(12)$ we get an isomorphic group. That $(45)$ part doesn't change the isomorphism type of the group because it commutes with $(123)$. But, it has the effect of turning $(12)$ (as well as the other odd permutations of $S_3$) into even permutations in $S_5$. – Jyrki Lahtonen May 11 '17 at 17:13
  • I guess a point is that $A_5$ has no cyclic subgroup of order six, so you have to look for a subgroup isomorphic to $S_3$ instead. – Jyrki Lahtonen May 11 '17 at 17:18
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    You can find a copy of $S_{n}$ in $A_{n+2}$ as follows: take an element of $S_n$. If it is already an even permutation, fix the extra two points. If it is an odd permutation, exchange the extra two points. This gives you an even permutation, hence an element of $A_{n+2}$. There are two orbits, and what happens to the final two points just keeps track of whether the permutation of the first $n$ is even or odd - the isomorphism is clear. – Mark Bennet May 11 '17 at 17:36

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https://groupprops.subwiki.org/wiki/Twisted_S3_in_A5

Read the above link, there is a subgroup of $A_5$ which is quite similar to $S_3$

Arpan1729
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