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Find all homomorphisms from $\mathbb Z_4\to\mathbb Z_2\oplus\mathbb Z_2$

Since $\mathbb Z_4$ is cyclic the homomorphisms are completely determined by the image of $1.$

Possibilities: $$1\mapsto(0,0)\\1\mapsto(0,1)\\1\mapsto(1,0)\\1\mapsto(1,1)$$

All four are homomorphisms, which can be found by direct calculation. E.g. if $\phi:1\mapsto(1,0)$

for $x,y\in \mathbb Z_4$: \begin{align*} \phi(x+y)&=(x+y \mod 2,0)\\&=((x \mod 2)+_2(y \mod 2),0)\\&=(x \mod 2,0)+(y \mod 2,0)\\&=\phi(x)+\phi(y). \end{align*}

Am I correct?

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  • $\begingroup$ I don't understand why you need to ask whether this is correct...you KNOW it's correct, you've checked exactly what you wanted/needed to check. $\endgroup$ – fretty Sep 26 '13 at 10:53
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This looks fine to me. You've checked all possible images of $1$ and have seen that all four possibilities give a homomorphism and so you have a full and correct list.

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The general way to find all homomorphism $\mathbb Z_n\to G$ for an arbitray abelian group $G$ is the following: Suppose $\phi:\mathbb Z_n\to G$ is a group homomorphism, as you said, it is determined by the image of $1$, so the question really is which choices of $g\in G$ give a homomorphism $\mathbb Z_n\to G$ when picked as the image of $1$?

For homomorphisms $\psi:\mathbb Z\to G$, this is easy: Pick any $g\in G$, let $\psi(x)=xg$ and just check $$\psi(x+y)=(x+y)g=xg+yg=\psi(x)+\psi(y).$$

Now what could go wrong? The subtle thing that happens here is better exposed, when we write $G$ as a multiplicative group. The definition of $\psi$ becomes $\psi(x)=g^x$ and the calculation becomes $$\psi(x+y)=g^{x+y}=g^x g^y =\psi(x)\psi(y).$$ The important fact: $g^x$ is defined for $x\in\mathbb Z$, that's why $\mathbb Z\to G$ is easy.

Going back to $\phi:\mathbb Z_n\to G$ and sticking to multiplicative notation, it is tempting to just choose $\phi(1)=g$ and define $\phi(x)=g^x$. But here you run into trouble: When $x\in\mathbb Z_n$, what is $g^x$ supposed to mean? Remember that $x\in\mathbb Z_n$ really is a coset of $n\mathbb Z$ in $\mathbb Z$, consisting of all elements of the form $x+kn$ with $k\in\mathbb Z$ and that $\mathbb Z_n$ is really just a shorthand for $\mathbb Z/n\mathbb Z$, the set of all cosets equipped with addition.

What you are really doing is defining $\psi:\mathbb Z\to G$ by $\psi(x)=g^x$ (which is fine) and then look at the induced map $\phi:\mathbb Z_n\to G$ with $\phi(x+n\mathbb Z)=g^x$ for all $x\in\mathbb Z$. Now the definition depends on the representative $x$ of the coset $x+n\mathbb Z$. For the map to be even well defined, it has to be independent of the choice of the representative: If $x+n\mathbb Z=y+n\mathbb Z$, so $x-y=kn$ for some $k\in \mathbb Z$, we want $g^x=g^y$, so $1=g^{x-y}=g^{kn}$. Thus, for $\phi$ to be well defined, we need $g^{kn}=1$ for all $k\in\mathbb Z$: The order of $g$ needs to be a divisor of $n$. When this is the case, the same calculation as above reveals that $\phi$ is not only well defined, but a well defined group homomorphism.

Now that we know, that we can pick any $g\in G$ with order diving $n$, the example $\mathbb Z_4\to\mathbb Z_2\oplus\mathbb Z_2$ is a quick one: All elements of $\mathbb Z_2\oplus \mathbb Z_2$ are of order $1$ or $2$, both are divisors of $4$, so we can pick any of them.

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Let $\phi(1) = a \implies \phi(x) = xa$, for any $x\in \mathbb{Z}_4$. $ \ \ \ \phi(x + y) = (x+y)a = xa + ya = \phi(x) + \phi(y)$.

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  • $\begingroup$ Careful there, if one insists that we need a ring homomorphism, then only one possibility is available. $a^2=\phi(1^2)=\phi(1)=a$ forces $a=(1,1)$. $\endgroup$ – user714630 Sep 26 '13 at 10:23
  • $\begingroup$ Given that the OP used the group-theory tag, I think we can assume they only want group homomorphisms. $\endgroup$ – Dan Rust Sep 26 '13 at 10:23
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    $\begingroup$ Even for homomorphism of abelian groups we have to be more careful. For example there are no nontrivial homomorphisms $\mathbb Z_4\to\mathbb Z$. Do you see where your calculation might fail? $\endgroup$ – Christoph Sep 26 '13 at 10:29
  • $\begingroup$ Nope. You mean that $\phi(1) \neq 1$? $\endgroup$ – CommutativeAlgebraStudent Sep 26 '13 at 10:34
  • $\begingroup$ Try $\phi:\mathbb Z_4\to\mathbb Z$ with $1\mapsto 1$. $\phi(2+2)=\phi(0)=0$, but $\phi(2)+\phi(2)=2+2=4$, and $4\neq 0$ in $\mathbb Z$. $\endgroup$ – Christoph Sep 26 '13 at 10:35

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