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I am asked to find all group homomorphisms from $Z/4Z$ to $Z/6Z$.

Let $f:Z/4Z \rightarrow Z/6Z$ be such a homomorphism.

By definition we have $f(1) = 1$ and therefore $f(0)=f(1 * 0) = f(1) * f(0) = f(1) * 0 = 0.$

Moreover $2 *3 = 6 = 2 (\text{mod 4})$ so $$f(2) * f(3) = f(2*3) = f(2),$$ which implies that $f(3) = 1$.

Is this sufficient? Are all homomorphisms described by the condition that $f(1) = 1 = f(3)$ and $f(0) =0$?

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  • $\begingroup$ Are we talking group homomorphisms or ring homomorphisms? $\endgroup$ – Gregory Grant Jul 24 '16 at 14:44
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    $\begingroup$ Then you don't need $f(1)=1$. $\endgroup$ – Gregory Grant Jul 24 '16 at 14:45
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    $\begingroup$ What about $f(0)=0$, $f(1)=3$, $f(2)=0$, $f(3)=3$ $\endgroup$ – Gregory Grant Jul 24 '16 at 14:47
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    $\begingroup$ Are you perhaps confusing the multiplicative and additive identities here? The operation is addition so you must have $f(0)=0$ by definition, not $f(1)=1$. $\endgroup$ – Gregory Grant Jul 24 '16 at 14:49
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    $\begingroup$ The confusion seems to be that you think the operation in your groups is multiplication. In fact, it is addition. Multiplication does not make either $Z/4Z$ or $Z/6Z$ into a group, but addition does. $\endgroup$ – Andreas Blass Jul 24 '16 at 14:53
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$\mathbb Z/n\mathbb Z$ is a group under addition modulo $n$; multiplication modulo $n$ is well defined, but does not make $\mathbb Z/n\mathbb Z$ a group for $n>1$ since $0$ does not have a multiplicative inverse.

Recall for an abelian group $A$, one denotes the operation by $+$, the identity element by $0$, and for $a \in A$ and $n \in \mathbb N$, $a^n$ becomes $na = a + a +\cdots + a$ $n$-times.

Say $f: \mathbb Z/4\mathbb Z \to \mathbb Z/6\mathbb Z$ is a group homomorphism. Since $f$ is a group homomorphism, $f(0) = 0$. Now since $\mathbb Z/4\mathbb Z$ is a cyclic group generated by $1 \pmod 4$, $f$ is completely determined by $f(1)$. Since $4 f(1) = f(1) + f(1) + f(1) + f(1) = f(1+ 1 + 1 +1) = f(4) = f(0) = 0$, we see that the (additive) order of $f(1)$ divides $4$. There is no element of order $4$ in $\mathbb Z/6\mathbb Z$, so the order of $f(1)$ can be $1$ or $2$. If it's 1, then $f(1) = 0$ and $f$ is the zero map; if it's 2, then $f(1) =3$.

So there are two homomorphisms.

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