Group Homomorphisms (some questions about what I can and can't do)

Question

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.

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

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

and

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

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?

Answer 1

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.

Answer 2

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.