Question

Set up equations and solve them to find the
unknown numbers in the following cases.
A). 18 added to 6 times of z gives -2

Answer 1

Answer:

Exercise. Describe all homomorphisms from Z24 to Z18.

Solution. We need some preliminary discussion before we try to define the homomorphisms

to try to figure out how they might look like. Let f : Z24 → Z18 be a homomorphism. For

any integer k ≥ 0 notice that

f(k mod 24) = f((1 mod 24) + · · · + (1 mod 24)) (added k times)

= f(1 mod 24) + · · · + f(1 mod 24)

= k · f(1 mod 24).

It follows that a homomorphism f is completely determined by the value f(1 mod 24). Write

f(1 mod 24) = n mod 18 where n is an integer such that 0 ≤ n ≤ 17. Notice that

24n mod 18 = 24f(1 mod 24)

= f(24 mod 24)

= f(0 mod 24)

= 0 mod 18.

since any homomorphism maps the identity to identity. So 18 must divide 24n, so 3 must

divide 4n, hence 3 must divide n. So n must be among 0, 3, 6, 9, 12, 15.

The above discussion suggests how we might want to define our homomorphisms. We find

that any such homomorphism must send (1 mod 24) to (3k mod 18) for some k = 0, 1, · · · , 5.

It follows that there are at most 6 possible homomorphisms from Z24 → Z18, given by

f0, f1, · · · , f5 where

fk(r mod 24) = 3kr mod 18 (1)

for k = 0, 1, · · · , 5. First we need to check that for each of this k, equation (1) gives a well

defined function from Z24 to Z18. To this end, let r, r′ ∈ Z such that r mod 24 = r

′ mod 24.

Then (r − r

′

) is a multiple of 24. So (3kr − 3kr′

) = 3k(r − r

′

) is a multiple of 3 × 24 = 72.

In particular, (3kr − 3kr′

) is a multiple of 18. It follows that (3kr mod 18) only depends on

the congruence class of r modulo 24. So indeed equation (1) gives us a well defined function

from Z24 to Z18. Now we verify that for any r, s ∈ Z, we have

fk(r mod 24 + s mod 24) = fk((r + s) mod 24) = 3k(r + s) mod 18

= 3kr mod 18 + 3ks mod 18

= fk(r mod 18) + fk(s mod 18).

So each fk is indeed a homomorphism from Z24 to Z18.

At the risk of repeating ourselves, we recall that if f : Z24 → Z18 is any homomorphism,

then f(1 mod 18) must be of the form (3k mod 18) for some k = 0, 1, · · · , 5 and that any

homomorphism f is determined completely by the value of f(1 mod 18). For k = 0, 1, · · · , 5,

we have constructed a homomorphism fk such that fk(1 mod 18) = 3k mod 18. It follows

that there are six homomorphisms from Z24 to Z18 and these are given by f0, f1, · · · , f5 as

defined in equation (1).

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