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Define $h\colon \mathbb{Z}/\sim ~16 \to \mathbb{Z}/\sim 24$ by $h([a]16) = [3a]24$

a. Prove $h$ is well defined.

b. Compute $h(a)$ where $a = \{[0]16, [3]16, [6]16\}$.

c. Compute $h^{-1}([10]24)$

Is the following correct:

a. $h$ is well defined since each for all $a$ $|h([a]16)|\equiv 1$.

b. $\{[0]24, [9]24, [18]24\}$

c. $\emptyset$

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b) and c) are right (but you should probably show your work if you're turning it in as homework) Part a) isn't quite as simple as what you have there. Ask yourself what $a$ represents, as opposed to $[a]16$. – Zarrax Dec 5 '10 at 20:56
For a. how about for h of all equivalence classes of Z/~16 there is one equivalence class in Z/~24? – Bradley Barrows Dec 5 '10 at 21:18

Your answers to (b) and (c) are correct.

For (a), you need to prove that if a=b mod 16, i.e. $a-b=16n $ for some integer $n$, then $3a=3b$ mod 24, i.e. $3a-3b=24m$ for some integer $m$. Can you do that?

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