# Injective homomorphism $(\mathbb{Z}_{12}, +) \to (\mathbb{Z}_{18}, +)$

$f:G \rightarrow H$ is a group homomorphism

$g\in G$ is an element of order $n$.

(a) Prove that $f (g)$ is a final order and that the order of the element $f (g)$ parts n.

(b) What is the order of the element $g^{-1}$? What is the order of the element $g ^ m$ for $m\in \Bbb N$?

(c) Whether there is any injective homomorphism $(\Bbb Z_{12}, +) \rightarrow (\Bbb Z_{18}, +)$?

What $(\Bbb Z_{12}; +) \rightarrow (\Bbb Z_{24}; +)$? If so, find example.

i did this:

a) $f(g)$- is final order and parts n $f(g)^n = f(e)=e$ $f(g)$ - is final order

b) $g^n = e \iff (g^-1)=e$ $g^{-1})^m=e$ for $m<n \implies g^m=e$

So how can i do second (b) and (c) THANKS

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What is a «final order»? – Mariano Suárez-Alvarez Jan 3 '13 at 17:52
Hi: I put in some TeX formatting into your question. Since it's very useful for this site, and very easy to pick up, I encourage you to take a look at how it's done by clicking the timestamp on the "edited" tag above. Good luck! – rschwieb Jan 3 '13 at 17:52
@MarianoSuárez-Alvarez: Sorry Mariano for calling you here, but did I something wrong below. I will be glad to know my defects. Sorry and thank you friend. :) – Babak S. Jan 3 '13 at 18:49

b) if $g$ is of order $n$, then the cyclic subgroup it generates is isomorphic to $\frac{\mathbb{Z}}{n\mathbb{Z}}$. The order of the class of $k$ in $\frac{\mathbb{Z}}{n\mathbb{Z}}$ is given by $\frac{k}{gcd(k,n)}$, so the order of $g^m$ is $\frac{m}{gcd(n,m)}$.
c) if there is an injective morphism $\varphi:\frac{\mathbb{Z}}{m\mathbb{Z}}\longrightarrow \frac{\mathbb{Z}}{m'\mathbb{Z}}$, then the image $\varphi(\frac{\mathbb{Z}}{m\mathbb{Z}})$ is a subgroup of $\frac{\mathbb{Z}}{m'\mathbb{Z}}$ which is isomorphic to $\frac{\mathbb{Z}}{m\mathbb{Z}}$. However by Lagrange's theorem the order of a subgroup of any group $G$ is a divisor of the order of $G$, hence in our case $m$ must divide $m'$. So there is no morphism with $m=12$ and $m'=18$. Now for $m'=24$ you can take the morphism $\bar{k}\mapsto \bar{2k}$.
$m'\mid m$ or $m\mid m'$?? $2\mathbb Z_4\subset\mathbb Z_2$. ;-) – Babak S. Jan 3 '13 at 18:05
Let $\mathbb Z_{12}=\langle g\rangle$, so $|g|=12$. In Group Theory we face to a simple theorem saying that in a finite group we always have $|g|=|g^{-1}|$, so $|g|=12$ as well.