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The following is a snippet from Gallian's Contemporary Abstract Algebra:

  • Chapter 10-Example 10:

We determine all homomorphisms from $\Bbb Z_{12}$ to $\Bbb Z_{30}$.

By

$\phi(g^n) = (\phi(g))^n$ for all $n\in\Bbb Z$",

such a homomorphism is completely specified by the image of $1$.

That is, if $1$ maps to $a$, then $x$ maps to $xa$.

By Lagrange’s Theorem and by " If $|g|$ is finite, then $|f(g)|$ divides $|g|$"

we require that $|a|$ divide both $12$ and $30.$

So, $|a|$ can be $1, 2, 3$, or $6$.

Thus, $a = 0, 15, 10, 20, 5,$ or $25.$

This gives us a list of candidates for the homomorphisms.

That each of these six possibilities yields an operation-preserving, well defined function can now be verified by direct calculations. [Note that $\gcd(12, 30) = 6.$ This is not a coincidence!]

Can someone please explain why $\gcd(12,30)=6$ isn't a coincidence? I mean to ask why $\gcd(12,30)$ equals the number of homomorphisms from $\Bbb Z_{12}$ to $\Bbb Z_{30}$.

Shaun
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Krishan
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2 Answers2

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I prove a general result here (of which yours is a special case):

Theorem: There are $\gcd(m,n)$ homomorphisms from $\mathbb Z_m$ to $\mathbb Z_n$.

Proof: As you've already noted, we just need to find the number of possible images of $1$ in $\mathbb Z_n$. Note that order of the image of an element is a factor of the order of the element (this is because, if $s\in\mathbb Z_m$ has order $k$, then $0=\varphi(sk) = \varphi(s)k$, therefore, $k$ divides order of $\varphi(s)$). Since the order of $1$ in $\mathbb Z_m$ is $m$, therefore, order of the image of $1$ has to be a factor of $m$. But at the same time, it also is a factor of $n$, by Lagrange's theorem. Therefore, the image of $1$ in $\mathbb Z_n$ has to be a common factor of both $m$ and $n$, which are ultimately the factors of $\gcd(m,n)$. Since there are $\phi(k)$ elements of order $k$ in $\mathbb Z_n$ (where $\phi(\cdot)$ is the Euler's totient function), therefore, the number of possible images of $1$, and hence the number of homomorphisms from $\mathbb Z_m$ to $\mathbb Z_n$ is given by $$\sum_{d|\gcd(m,n)}\phi(d) = \boxed{\gcd(m,n)}$$

Shaun
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Martund
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GCD(12,30)=6,the factor of 6 are {1,2,3,6), i.e., you can have smaller subgroup of codomain Group ${Z_{30}}$ only of above order. That is what the property you mentioned above.

  • In Homomorphism if we sacrifice bijection requirement of Isomorphism, then we can have onto relation with codomain group of a domain Group.
  • GCD confirms that any of divisor of two group below its value, is a candidate for homomorphism, considering a important order related property described above.
  • Same Order of mapped elements is a required property, but not a sufficient property. It still needs to obey the ${\phi(ab)=\phi(a)\phi(b)}$
SSA
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