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Homework Help: Why doesn't the image of a group have the same cardinality as the group?

  1. Mar 26, 2012 #1
    I was doing one of the proofs for my abstract algebra class, and we had to prove that the cardinality of the image of G, [θ(G)] is a divisor lGl. I'm trying to intuitively understand why G and it's image don't necessarily have the same cardinality. I'm thinking it's because there isn't necessarily a one to one correspondence, if there is then they have the same cardinality, but if they don't then the cardinality of the image is less than the cardinality of G, am I thinking along the right track here? This isn't a homework problem, but I came across it in my work. Any insight is appreciated.

  2. jcsd
  3. Mar 26, 2012 #2


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    The image of G in what function?
  4. Mar 26, 2012 #3
    @HallsofIvy: Sorry, should have been more clear. For θ:G → G' and it's a group homomorphism. I was writing the proof for it, and was wondering.
  5. Mar 26, 2012 #4
    What are you allowed to use? If you know that the image of a homomorphism is a subgroup and you know the order of a subgroup divides the order of the group, you're done. Otherwise you'll have to prove one or both of those facts from first principles.
  6. Mar 26, 2012 #5
    @SteveL27: The image would be a subgroup of G', right, not G? I was just wondering, why the image of G and G itself don't necessarily have the same cardinality? I guessed this was because the map from G to G' is not necessarily one-to-one and onto, but not sure if i'm correct.
  7. Mar 26, 2012 #6
    Yes of course. Sorry, little brain hiccup at my end. Nevermind what I wrote. You're right, if the homomorphism's not onto then the image will be a proper subset of G'.

    (edit) Ok I owe you a better hint. Do you know the first isomorphism theorem?
    Last edited: Mar 26, 2012
  8. Mar 27, 2012 #7
    We recently covered that actually, and considering this is a hint, i'm going to look into it and see if it clarifies things. Thanks!
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