There is a unique group homomorphism from ZZ to any group where the image of 1 is fixed

Show that if H is a group and h \in H, there exists a unique homomorphism \varphi : \mathbb{Z} \rightarrow H such that \varphi(1) = h.

Existence: Define \varphi(n) = h^n. We need not worry about well-definedness for \varphi. For all m,n \in \mathbb{Z}, we have \varphi(m+n) = h^{m+n} = h^mh^n = \varphi(m)\varphi(n), so that \varphi is a homomorphism.

Uniqueness: Suppose we have another homomorphism \psi such that \psi(1) = h. Then \psi(n) = \psi(n \cdot 1) = \psi(1)^n = \varphi(1)^n = \varphi(n), so that \psi = \varphi and so \varphi is unique. (Keep in mind that we write Z additively and H multiplicatively.)

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  • Bobby Brown  On September 14, 2010 at 1:54 pm

    In this one, I am having trouble understanding why:


    can anyone give a bit more explanation to this step?

    • nbloomf  On September 14, 2010 at 2:11 pm

      It is because we’re thinking of \mathbb{Z} as an additive group. There, the identity is 0 and “powers” are written as “multiples”. H, on the other hand, is written multiplicatively. I can see why this might be confusing- I’ll add a bit of explanation.

      (Thanks for reading, by the way!)

  • Bobby Brown  On September 14, 2010 at 4:24 pm

    No I understand that part. Ah I see now though…you meant psi(1+1+…+1), with 1 appearing n times, then you can split that into psi(1)*…*psi(1), also appearing n times, and so on. Thanks and you’re welcome.

    I study math at Colorado (I was accepted to Arkansas, but opted for here…) and this has been a great algebra resource for me. Do you know of something similar to this in analysis, perhaps with Royden’s book?

    Good luck with your studies

    • nbloomf  On September 14, 2010 at 7:35 pm

      I’m glad to hear you find this useful. Unfortunately I don’t know of any similar resources for analysis.

      Good luck to you as well.

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