# Question about generators and relations.

#### Jack

##### New member
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

##### Well-known member
MHB Math Helper
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?

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#### Deveno

##### Well-known member
MHB Math Scholar
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

##### Well-known member
MHB Math Helper
perhaps you meant to use "M", instead of "N",
Of course, just a typo.