Question about generators and relations.

[Jack3]

I am trying to use generators and relations here.

Let M ≤ S_5 be the subgroup generated by two transpositions t_1= (12) and t_2= (34).

Let N = {g ∈S_5| gMg^(-1) = M} be the normalizer of M in S_5.

Describe N by generators and relations.

Show that N is a semidirect product of two Abelian groups.

Compute |N|.

How many subgroups conjugate to M are there in S_5 ? Why？

(I think Sylow's theorems should be used here.)

 

[Fernando Revilla]

Describe N by generators and relations.

Please, show some work, $\color{red}M\color{black}=\{(1,2)^m(3,4)^n:m,n\in \mathbb{Z}\}$. What do you obtain?

 

[Deveno]

Please, show some work, $N=\{(1,2)^m(3,4)^n:m,n\in \mathbb{Z}\}$. What do you obtain?

are you sure about this? it seems to me that g = (1 3)(2 4) is an element of N, since:

gt₁g^-1 = t₂

gt₂g^-1 = t₁

g(t₁t₂)g^-1 = t₂t₁ = t₁t₂ (since these are disjoint, and thus commute).

perhaps you meant to use "M", instead of "N", N is the normalizer of M, and we might expect to to be a bit larger than M itself (of course it contains M as a subgroup).

i claim it is obvious that |M| = 4, and that M is non-cyclic. i also claim that no element of N can move 5. so |N| is between 4 and 24, and is a multiple of 4. you should prove these things.

this gives 4 possibilities: |N| = 4,8,12, or 24. since i show an element of N not in M above, 4 is off the table. it can be shown by direct computation that:

t₁t₂g = (1 4)(2 3) is also in N. this gives a second subgroup of N of order 4:

A = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}.

some things for YOU to do: show N doesn't contain any 3-cycles. this means |N| cannot be 24 OR 12 (since the only subgroup of order 12 of S₄ is A₄ which contains ALL 3-cycles).

so |N| = 8, and furthermore N has at least 5 elements of order 2. which group of order 8 could this be?

abelian possibilities:

Z₈ (has only one element of order 2), Z₄xZ₂ (has 3 elements of order 2), Z₂xZ₂xZ₂ (has 7 elements of order 2).

if it turned out N had an element of order 4, it must be non-abelian. does it?

something that may or may not be relevant: Q₈ has 6 elements of order 4, and only 1 element of order 2.

finally, if you arrive at the right choice for N, i hope you will clearly see there is an easy way to see it as a semi-direct product of abelian groups (hint: it has a normal subgroup of index 2).

 

[Fernando Revilla]

perhaps you meant to use "M", instead of "N",

Of course, just a typo.

 

[blue_raver22]

I would first clarify the context of the question and make sure I understand the notations and definitions being used. From my understanding, S_5 refers to the symmetric group of order 5, which consists of all permutations of 5 elements. M is a subgroup of S_5 generated by two specific transpositions, and N is the normalizer of M in S_5, which consists of all elements in S_5 that commute with M.

To describe N by generators and relations, we can use the fact that N is generated by the set of all elements in S_5 that are conjugate to M. In this case, we have two generators t_1 and t_2, which correspond to the transpositions (12) and (34), respectively. As for relations, we can use the fact that t_1 and t_2 are both of order 2 and commute with each other, so t_1^2 = t_2^2 = (t_1t_2)^2 = e, where e is the identity element in S_5.

To show that N is a semidirect product of two Abelian groups, we can use the fact that N is generated by two commuting elements t_1 and t_2. This implies that N is Abelian, as any two elements in N commute with each other. Furthermore, since t_1 and t_2 generate N, we can write N as the direct product of the cyclic subgroups generated by t_1 and t_2. Since both of these subgroups are Abelian, N must also be Abelian.

To compute |N|, we can use the fact that |N| = |S_5|/|M|, where |S_5| = 5! = 120 and |M| = 2. Therefore, |N| = 120/2 = 60.

Using Sylow's theorems, we can determine the number of subgroups conjugate to M in S_5. Since M is a subgroup of order 2 in S_5, by Sylow's third theorem, the number of subgroups conjugate to M must divide 5!. However, since M is not a normal subgroup of S_5, the number of subgroups conjugate to M cannot be 5! itself. Therefore, by Sylow's second theorem, the number