1
$\begingroup$

I am having a hard time studying and I am a visual learner. How could I visually imagine a (group) homomorphism $$\mathbb{Z}_{12} \to \mathbb{Z}_3?$$

Also, if the question states that the map $f$ is a group homomorphism such that $f(1)=2$, how could I find the kernel $K$ of $f$?

$\endgroup$
  • 2
    $\begingroup$ If $f(1)=2$ then $f(1+1)=2+2=1$ and $f(1+1+1)=0$. What else gets mapped to $0$? $\endgroup$ – André Nicolas Feb 4 '15 at 5:35
  • 1
    $\begingroup$ Supposing that by $\Bbb Z_n$ you mean the integers modulo $n$, there are $3^{12}=531441$ different such maps. They all look different, so one cannot answer this question unless you indicate what map you mean. You later say "homomorphism" but that depends on what kind of structure you are giving to $\Bbb Z/n\Bbb Z$: group, ring, other? The tag "abstract-algebra" is not helpful either here. In fact no ring homomorphism satisfies $f(1)=2$, so probably you mean homomorphism of additive groups $\endgroup$ – Marc van Leeuwen Feb 4 '15 at 5:35
  • $\begingroup$ It was a question on my textbook under section Isomorphism Theorems (Advanced Group Theory) $\endgroup$ – Lilmaymay4 Feb 4 '15 at 5:43
  • $\begingroup$ @Lilmaymay4 I've rewritten the question to be a little more specific (and to address Marc's well-placed comments). Please adjust/revert the changes if I misrepresented your intention. $\endgroup$ – Travis Willse Feb 4 '15 at 5:58
  • $\begingroup$ @MarcvanLeeuwen Aw, come on. $\endgroup$ – Pedro Tamaroff Feb 4 '15 at 6:01
2
$\begingroup$

I find it useful to imagine the group $\mathbb{Z}_n$ a group of rotations (of, e.g., the plane) by multiples of $\frac{1}{n}$ revolution, that is, by multiples of $\frac{2\pi}{n}$ radians, so that $[k] \in \mathbb{Z}_n$ corresponds to a (say, anticlockwise) rotation by $\frac{2 \pi k}{n}$ radians.

Since $\mathbb{Z}_n$ is cyclic, any group homomorphism $f: \mathbb{Z}_n \to H$ is determined by $f(1)$. This leaves three candidate maps, namely - the map defined by $f([1]) = [0]$, which is just the zero homomorphism $f([n]) := 0$, - the map defined by $f([1]) = [1]$, - the map defined by $f([1]) = [2]$.

Per the above mnemonic, we can regard $[1] \in \mathbb{Z}_{12}$ as an anticlockwise rotation by $\frac{1}{12}$ of a revolution, or $\frac{\pi}{6}$ radians, and, e.g., $[2] \in \mathbb{Z}_{3}$ as an anticlockwise rotation by $\frac{2}{3}$ of a revolution, or $\frac{4\pi}{3}$ radians. So, using that $f$ is a group homomorphism, so that $f([n]) = n \cdot f([1])$ we can think of $\phi$ as the map that takes a given rotation in $\mathbb{Z}_{12}$ and applies it $\frac{\frac{2}{3} \text{ rev}}{\frac{1}{12} \text{ rev}} = 8$ times, which by construction is always a rotation in $\mathbb{Z}_3$ so imagined.

(As a word of warning, this is a convenient way to think about finite cyclic groups, but if one wants to think about $\mathbb{Z}_3$ sitting inside $\mathbb{Z}_{12}$ the way we did here, one must specify that one is using this mnemonic; indeed, there is more than one way to put $\mathbb{Z}_3$ into $\mathbb{Z}_{12}$ in a way that respects group multiplication.)

Thinking of a group as we did here as a set of (linear) transformations of some vector space, by the way, is a (the) central idea of representation theory, which has proved to be an immensely powerful too for understanding groups.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.