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Define a function $f_n:\mathbb{Z}_m \to \mathbb{Z}_m$ as a map $\bar{a}\mapsto n\cdot \bar{a}$. Show that it is both well-defined and a group homomorphism.

For the well-defined part, I know that I need to somehow show that when $a_1=b_1 \in \mathbb{Z}_m$, then $f(a_1)=f(b_1)\in \mathbb{Z}_m$. I just don't know where to start. For the homomorphism, $f(a_1b_1)=f(a_1)f(b_1)$, I'm also unclear as how to proceed.

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  • $\begingroup$ Does the homomorphism hinge on $\bar{n}\in \mathbb{Z}\_m$? $\endgroup$ – beethree Feb 19 '13 at 21:57
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Assume $a_1=b_1\,\in\Bbb Z_m$. What does it mean? That $a_1\equiv b_1 \pmod m$, i.e. $m|\,b_1-a_1$. Then $f(b_1)=[n\cdot b_1]$ where $[x]$ denotes the equivalence class of $x$ w.r.t. mod $m$... You will have to prove that $[n\cdot a_1]=[n\cdot b_1]$.

Similarly, playing with the equivalence classes by arbitrary representants will help you in the other part.

Note that a homomorphic relation $R$ between (additively written) groups $A$ and $B$, such that, for all $a$ there is $b$ such that $a$ and $b$ are in the relation $R$ --written as $aRb$--, is actually a function (homomorphism) iff $0Rb\Rightarrow b=0$, i.e. it is enough to verify that $0$ can go only to $0$.

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  • $\begingroup$ is it true that $n\cdot [b_1]=[n\cdot b_1]$? Couldn't I just say that $f(b_1)=n\cdot [b_1]=n\cdot [a_1]=f(a_1)$? $\endgroup$ – beethree Feb 19 '13 at 20:42
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Who on earth has created this "exercise"?

The used notation $n \cdot \overline{a}$ means that well-definedness is already assumed! It would be different for $\overline{n \cdot a}$, but this expression doesn't appear in the statement. Besides: This has nothing to do with $\mathbb{Z}_m$ (which seems to be the usual wrong/ambigious notation for $\mathbb{Z}/m\mathbb{Z}$).

If $A$ is an arbitrary abelian group, and $n \in \mathbb{N}$, then $A \to A, a \mapsto n \cdot a$ is a homomorphism. Hint for the proof: Recall the inductive definition of $n \cdot a$. Show $n \cdot (a+b)=n \cdot a + n \cdot b$ by induction on $n$.

One gains nothing and it doesn't get easier by restricting to this artificial special case $A=\mathbb{Z}/m\mathbb{Z}$.

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Here is a rather long argument for what you are trying to prove, but I thought it could perhaps be interesting to see that your homomorphism arises in a very natural manner.

First note that $$ a\longmapsto na $$ is a homomorphism from $\mathbb{Z}$ to itself.

Then compose with the canonical surjection of $\mathbb{Z}$ onto $\mathbb{Z}/m\mathbb{Z}$ to get a homomorphism $$ a\longmapsto \overline{na}=n\overline{a} $$ from $\mathbb{Z}$ to $\mathbb{Z}/m\mathbb{Z}$.

Finally observe that $m\mathbb{Z}$ is contained in the kernel of the latter, so that you homomorphism factors through the quotient and leads to a homomorphism $$ \overline{a}\longmapsto n\overline{a} $$ from $\mathbb{Z}/m\mathbb{Z}$ into itself.

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  • $\begingroup$ This is a touch beyond where we are at currently in the class. Still, though, thank you. $\endgroup$ – beethree Feb 18 '13 at 22:51
  • $\begingroup$ @beethree You're welcome. $\endgroup$ – Julien Feb 19 '13 at 2:03

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