Determine all pairs n, m elementof N such that there exists a non-trivial homo-m

Question

Determine all pairs n, m elementof N such that there exists a non-trivial homo-morphism from Z/nZ to Z/mZ. Does [a](s) = a + s define an action of (Z/nZ, +) on R. Define an action of Sym_3 on R^3 as follows: for sigma elementof Sym_3 and upsilon = (x_1, x_2, x_3) elementof R^3, set sigma middot upsilon = (x_sigma(1), x_sigma(2), x_sigma(3)). Show that the subspace V = {(x_1, x_2, x_3)|x_1 + x_2 + x_3 = 0} has the following property: if upsilon elementof V, then sigma middot upsilon is also in V. Can you find a line (one-dimensional subspace) which has this same property?

Explanation / Answer

Ans 9
If m and n are relatively prime, we can explicitly display an isomorphism.

Define f: Z --> Zn x Zm by f(k) = (k mod n, k mod m).
It is easy to check that this yields a ring homomorphism.

ker f = {k in Z : f(k) = (0 mod n, 0 mod m)}
= {k in Z : n|k and m|k}
= {k in Z : mn | k}, since m and n are relatively prime
= (mn)Z.

The only 'tricky' part is showing that f is surjective.
Given (a, b) in Zn x Zm, we need to find k in Z such that f(k) = (a, b).
Since m and n are relatively prime, there exist x, y in Z such that mx + ny = 1
Note that mx = 1 (mod n) and ny = 1 (mod m).

Let k = a * (mx) + b * (ny). Then f(k) = (a * 1 + 0, 0 + b * 1) = (a, b) as required.
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Finally, the first isomorphism theorem yields the desired result.

According to the Chinese Remainder Theorem, if M and n are relatively prime, there is an isomorphism of rings:Zmn=Zm×Zn.

Then using part (a) and the definition of given above:

(mn) =|Zmn|=|(Zm×Zn)|=|Zm×Zn|=|Zm||Zn=(m)(n)

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