1 let dn denote the dihedral group the group of symmetries...

1. Let $D_n$ denote the Dihedral group (the group of symmetries of a regular $n$-gon), and let $S_n$ be symmetric group (the group of bijections on $X_n = \{1,...,n\}$). Are $D_3$ and $S_3$ isomorphic? Justify your answer. Note that both groups have order 6. Bonus Question Prove or disprove: $D_{12} \cong S_4$.

Transcript

00:01 In this problem, our first job is to find the least positive integer m, such that z12 is isomorphic to a subgroup of the symmetric group s sub m.

00:09 And then, to determine if we know we have a permutation s sub m that has order 12, must that permutation necessarily be even or odd? so we will start, as the problem notes, by recalling how to calculate the order of a permutation in the group s sub m.

00:28 So the order of a permutation, we'll call it f, is the least common multiple of the length of its disjoint cycles.

00:43 Those cycles have to be disjoint.

00:51 So that allows us to calculate the order of any element in s.

00:56 So for example, we might consider when m is 6, we would ask the question, what is the maximum order an element can have in s ? 6.

01:13 Would 6 be big enough to allow us to find a subgroup of s6 that is isomorphic to z12? well, there are different ways we can partition 6.

01:29 We can just have a 6 cycle.

01:32 So if we have a 6 cycle in s6, we know its order is going to be 6.

01:43 We can try to partition 6 in different ways.

01:46 We can partition a 5 at 1, but that's going to give an order of 5 using the least common multiple property.

01:52 If we have a product of a three cycle and a two cycle, for example, we could take the permutation 1, 2, 3, 4, 5 in s6, but its order – we'll use little o for order – would be the least common multiple of 3 and 2, and that would be 6.

02:14 So it turns out we can't improve on 6 in s6.

02:17 That really is the maximum order of an element in s6.

02:21 So s6 is not going to work.

02:24 M equals 6 is not big enough.

02:27 So what we're claiming here, just looking at a couple of examples, is that in s6 the maximum order of an element is 6.

02:40 That's the best we can do.

02:43 But if we just go up one more to m equals 7, 7, we can achieve elements of order 12 in s7.

02:53 So when m is 7, then we can consider a permutation that is the product of a 4 -cycle and a 3 -cycle.

03:05 We can just do, of course, disjoint cycles.

03:08 We can let f be permutation 1, 2, 3, that 3 -cycle composed with the 4 -cycle 4, 5, 6, 7.

03:16 So we know that the order of f is the least common multiple of the 3 and the 4, so we have achieved an order of 12.

03:30 So that means that the cyclic subgroup generated by f in s7 has order 12, and it's cyclic by definition.

03:51 So therefore, since cyclic groups of the same in order are isomorphic, we know that z12 is isomorphic to – we use isomorphic, that notation there is z12 is isomorphic to the cyclic subgroup generated by this permutation f, and f, remember, is an s7.

04:13 So our answer for part one is that we can get away with m equals 7, the smallest positive positive integer for which we will have a subgroup of s of m that is isomorphic to z12...

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