Here is an example on Hungerford's Algebra, page 31. enter image description here

But I don't think this argument is correct, according to the definition of homomorphism, $g(xy)=g(x)g(y)~x,y\in \mathbf{Z}_m$ that means the congruence equation $$ k(xy\bmod{m}) \equiv (kx)(ky)\pmod{km}$$but the preceding equation may not be true for an arbitrary positive integer $k$.

Did I have any misunderstanding on this example?

  • 4
    $\begingroup$ The operation is addition, not multiplication. $\endgroup$ Sep 19, 2016 at 10:34
  • $\begingroup$ Just to emphasize: $\mathbb Z_m$ is not even a group under multiplication; $0$ has no inverse. $\endgroup$
    – lulu
    Sep 19, 2016 at 10:36

1 Answer 1


$Z_m$ is not always a group under multiplication. So when he is referring to the map

$g: Z_m \to Z_{mk} $ $x \mapsto kx$

then $g(x+y) = k(x+y)=kx+ky=g(x)+g(y)$

thus it's a homomorphism. As to why is injectiv just note, that the image has the same number of elements as $Z_m$ has.


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