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Using the Second Isomorphism (Diamond Isomorphism) Theorem

  1. Oct 2, 2015 #1
    1. The problem statement, all variables and given/known data
    Good day all,

    Im completely stumped on how to show this:

    |AN|=(|A||N|/A intersect N|)

    Here: A and N are subgroups in G and N is a normal subgroup.
    I denote the order on N by |N|

    2. Relevant equations

    Second Isomorphism Theorem


    3. The attempt at a solution

    Well, I know know im supposed to use the theorem, but I have no idea where to start!
    I dont seem how to relate what the theorem says to orders of sets....

    Any help would be greatly appreciated!
     
  2. jcsd
  3. Oct 2, 2015 #2

    andrewkirk

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    First, to be able to use the 2nd isomorphism theorem at all, you need to prove that AN is a subgroup. Has that been given to you as a theorem? If not, can you prove it (it's quite easy)?

    Next note that the problem statement only makes sense if A and N are finite.

    With those out of the way, the second isomorphism theorem says:

    $$AN / N\cong A/A\cap N$$
    from which it follows that
    $$|AN / N|=| A/A\cap N|$$
    You have been asked to prove

    $$\frac{|AN|}{|N|}=\frac{|A|}{|A\cap N|}$$

    So if you can prove that, for any finite group ##G## and normal subgroup ##N##,
    $$|G/N| =\frac{|G|}{|N|}$$
    then you're almost done.

    Can you prove that? Think about the size ##|gN|## of each coset ##gN##.
     
  4. Oct 2, 2015 #3
    Ahh! Thanks! I didnt realize I could just divide by |N|....I am new to quotient groups and for some reason I havnen't been able to get the hang of them yet. Thanks!

    I'll use Lagranges Theorem for the last part
     
  5. Oct 2, 2015 #4

    andrewkirk

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    You can do that because the equation you've written is just numbers, as everything is enclosed by |.| signs. So no group theory is needed to justify the division.

    Actually, the division is only justified if ##|N|\neq 0## but the proposition is untrue otherwise, since the LHS will be 0 and the RHS will be 0/0, which is undefined. So we assume that the examiner meant to stipulate that ##|N|\neq 0##.
     
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