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Order of groups in relation to the First Isomorphism Theorem.

  1. Nov 14, 2008 #1
    Given H,K and general finite subgroups of G,

    ord(HK) = [(ord(H))(ord(K))] / ord(H intersection K)

    I know by the first isomorphism theorem that Isomorphic groups have the same order, but the left hand side of the equation is not a group is it?

    I am struggling to show this.
  2. jcsd
  3. Nov 14, 2008 #2


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    Yes, HK is not necessarily a group, but this is irrelevant. The identity you posted follows from an easy counting argument. The only theorem you need is Lagrange's. Here's a hint: [itex]HK = \cup_{h \in H} hK[/itex]. So ord(HK) = ord(K) * number of distinct cosets of K of the form hK.
  4. Nov 21, 2008 #3
    This doesn't help your problem, but if one of the subgroups, say H, is normal in G, then HK is a subgroup of G and (HK)/H and K/(H ∩ K) are isomorphic (this is the so-called second isomorphism theorem), from which the statement about the orders follows easily.
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