Subgroup of A4: Even Permutations on 4 Elements

[Punkyc7]

let M={x$^{2}$|x$\in$G} where G is a group.
Show M is not a sub group if G=A$_{4}$(even permutations on 4 elements.)

(Since an even times an even is even its closed, the identity is even and all the inverses should be even.)* Is is associativity were it is going to go wrong? Or am I wrong in thinking *.

 

[micromass]

Can you actually calculate M for me?? That is: can you tell me what the elements of M are?

 

[Punkyc7]

e^2=e
(123)^2=(132)
(124)^2=(142)
(134)^2=(143)
(234)^2=(243)
(132)^2=(123)
(142)^2=(124)
(143)^2=(134)
(243)^2=(234)
((12)(34))^2=e
((13)(24))^2=e
((14)(23))^2=e

I think I got them all there should be 12

 

[Deveno]

e^2=e
(123)^2=(132)
(124)^2=(142)
(134)^2=(143)
(234)^2=(243)
(132)^2=(123)
(142)^2=(124)
(143)^2=(134)
(243)^2=(234)
((12)(34))^2=e
((13)(24))^2=e
((14)(23))^2=e

I think I got them all there should be 12

ok, so you know that every square of an element of A₄ is again in A₄. why is this not enough to show closure?

hint: what is (1 2 4)(1 2 3) ? is that a square of something in A₄?

 

[Punkyc7]

I want to show M is not a subgroup, but I can't seem to figure out were it is going to go wrong. All even permutations squared are even. Their inverse exist because they are also even and there is an identity.

 

[Deveno]

yes M is a subSET of A₄. is it a subGROUP?

look, i'll give an analogy:

consider the set of even integers under addition. that is a group, right? now consider the set:

{k in Z: k = (2n)², n in Z}.

well, all these integers are still even, but they do not form a subgroup of the even integers. why? because even though 4 and 16 are in this set, 20 = 4 + 16 is NOT.

you have to show MORE than just that the resulting squares are even permutations. you have to show that every PRODUCT of squares of even permutations is ALSO a square of even permutations. ask yourself: is the product of 2 3-cycles always another 3-cycle, or the identity?

 

[Punkyc7]

im not quite sure what you are trying to say. Is it because if you multiply any of the above won't you get an even permutation? The product of the identity is always the identity and the product of a 3 cycle I am not sure about, I am guessing maybe not?

 

[Joffan]

Definition of a subgroup H of G:

• H is closed under the group operation of G,
• the identity element of G is in H and
• for every a in H, a^-1 is also in H.

Have you shown all these properties, or found a violation of one of them?

In fact, since this is a finite group, checking the first property is enough.

Incidentally, [cough] Lagrange's Theorem

Spoiler

The order of a subgroup H of group G divides the order of G

 

[Punkyc7]

is it because (132)(142)=(14)(23) and (14)(23) isn't a square?

 

[Joffan]

(14)(23) isn't a member of M. Therefore...

 

[Punkyc7]

not closed...cant believe I didn't see that.