Let $G$ be a group and fix some $g\in G$. Consider the map $\phi:\mathbb{Z}\rightarrow G$ defined by $\phi(n)=g^n$. Show that $\phi$ is a group homomorphism.

Let $a,b\in \mathbb{Z}$. Then,

$\hspace{150pt} \phi(a+b)=g^{a+b}$


$\hspace{150pt} \phi(a)+\phi(b)=g^ag^b=g^{a+b}$

Did I do this correctly? $\mathbb{Z}$ is a group under addition, but this doesn't work if I make $G$ a group under addition too. Am I allowed to let $G$ be a group under multiplication instead?

  • $\begingroup$ it should be: phi(a)*phi(b) and not + $\endgroup$ – DeepSea Mar 6 '14 at 4:33
  • $\begingroup$ But then I would have to have $\phi(a*b)$ as well. $\endgroup$ – TheMobiusLoops Mar 6 '14 at 4:34
  • $\begingroup$ If your group is abelian, you would want the map $\phi$ to send $n$ to $ng$ in order to get a group homomorphism. If your group operation is addition, you have no notion of multiplying group elements, unless, say, you are looking at the additive group of a ring. $\endgroup$ – Chris Leary Mar 6 '14 at 4:40

Note that ${\Bbb Z}$ is a specific group in which the operation is addition. But $G$ in your question is an unspecified group in which the operation could be anything (as long as the group axioms are satisfied). You can write the operation as multiplication, in which case $g\,g\,\cdots\,g$ with $n$ factors will be abbreviated to $g^n$. But this doesn't mean that the operation actually is multiplication, and it certainly doesn't mean you have to take the same operation for ${\Bbb Z}$. If you wished, you could write the operation in $G$ as addition, in which case the question would probably be stated this way:

Consider the map $\phi:{\Bbb Z}\to G$ defined by $\phi(n)=ng$.

Remember again, this doesn't mean the operation in $G$ really is addition, it just means it's written as addition. More abstractly (but perhaps more clearly), you could let the operation be an unspecified $*$, writing the question as

Consider the map $\phi:{\Bbb Z}\to G$ defined by $\phi(n)=g*g*\cdots*g$, where there are $n$ factors of $g$ on the right hand side,

although this would have the drawback that if $n$ is negative it's not entirely clear what the right hand side means.

Hope this helps.


The logic here looks good (if you're looking for a proof I would make it a bit more linearly - comment down below if you want me to explain further on this).

Note that in a group there is only one operation, let's denote the operation on $G$ as $\circledast$. Now it is natural to denote $$ g^n = \underbrace{g \circledast g \circledast \ldots \circledast g}_{n \text{ times}} $$ similar to how in $(\mathbb{R} \setminus \{ 0 \}, \cdot)$ we have $$ r^n = \underbrace{r \cdot r \cdot \ldots \cdot r}_{n \text{ times}} $$

The key thing to see in your group $G$, you can call the operation what ever you want, addition, multiplication, truncation, elephant-ification, nonetheless the notation above still holds.

  • $\begingroup$ So you would have me write it as $\phi(a+b)=g^{a+b}=g^ag^b=\phi(a)+\phi(b)$, right? $\endgroup$ – TheMobiusLoops Mar 6 '14 at 4:53
  • $\begingroup$ @TheMobiusLoops yep exactly! Hopefully this helps! $\endgroup$ – DanZimm Mar 6 '14 at 4:54
  • $\begingroup$ I just want to make sure I have shown that this is a homomorphism. I have a test coming up, and want to make sure things are straight. $\endgroup$ – TheMobiusLoops Mar 6 '14 at 4:56
  • 1
    $\begingroup$ @TheMobiusLoops Well implicitly there is one thing that you're missing: $\phi$ needs to be a function, in other words it needs to be well defined. So here I would say something along the lines of, if $n=m$ then it's clear that $g^n = g^m \iff \phi(n) = \phi(m)$. This is simple in this example because you can't represent elements in $\mathbb{Z}$ multiple ways but in for example $\mathbb{Z} / 5$ we have $n = n + 5k$ for any integer k. $\endgroup$ – DanZimm Mar 6 '14 at 5:01
  • $\begingroup$ I see, thank you. You have been very helpful. $\endgroup$ – TheMobiusLoops Mar 6 '14 at 5:03

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