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Homework Help: Subgroup Question

  1. Jan 21, 2012 #1
    let M={x[itex]^{2}[/itex]|x[itex]\in[/itex]G} where G is a group.
    Show M is not a sub group if G=A[itex]_{4}[/itex](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 *.
  2. jcsd
  3. Jan 21, 2012 #2
    Can you actually calculate M for me?? That is: can you tell me what the elements of M are?
  4. Jan 21, 2012 #3

    I think I got them all there should be 12
  5. Jan 21, 2012 #4


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    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?
  6. Jan 21, 2012 #5
    I want to show M is not a subgroup, but I cant 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.
  7. Jan 21, 2012 #6


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    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?
  8. Jan 21, 2012 #7
    im not quite sure what you are trying to say. Is it because if you multiply any of the above wont you get an even permutation? The product of the identity is always the identity and the product of a 3 cycle im not sure about, im guessing maybe not?
  9. Jan 21, 2012 #8
    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
    The order of a subgroup H of group G divides the order of G
  10. Jan 21, 2012 #9
    is it because (132)(142)=(14)(23) and (14)(23) isnt a square?
  11. Jan 21, 2012 #10
    (14)(23) isn't a member of M. Therefore....
  12. Jan 21, 2012 #11
    not closed.....cant believe I didn't see that.
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