Let φ: Z/24Z→Z/16Z be a group homomorphism. What are the possibilities for the image of φ and the corresponding possibilities for ker(φ)?
Since φ is completely determined by where its sends a generator g of Z/24Z, we have
1 = φ(g^24) = (φ(g))^24. So, the order of φ(g) is a divisor of 24.
On the other hand, |φ(g)| must be a divisor of |Im φ|, which divides |Z/16Z| = 16.
==> |φ(g)| = 1, 2, 4, 8, or 16.
Putting this all together, we deduce that |φ(g)| = 1, 2, 4, or 8.
This leaves four possibilities (keeping in mind that (Z/24Z) / ker φ ≅ Im φ and so 24/|ker φ| = |Im φ|):
|φ(g)| = 1 ==> |Im φ| = 1 and thus |ker φ| = 24/1 = 24.
|φ(g)| = 2 ==> |Im φ| = 2 and thus |ker φ| = 24/2 = 12.
|φ(g)| = 4 ==> |Im φ| = 4 and thus |ker φ| = 24/4 = 6.
|φ(g)| = 8 ==> |Im φ| = 8 and thus |ker φ| = 24/8 = 3.
I hope this helps!
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Since φ is completely determined by where its sends a generator g of Z/24Z, we have
1 = φ(g^24) = (φ(g))^24. So, the order of φ(g) is a divisor of 24.
On the other hand, |φ(g)| must be a divisor of |Im φ|, which divides |Z/16Z| = 16.
==> |φ(g)| = 1, 2, 4, 8, or 16.
Putting this all together, we deduce that |φ(g)| = 1, 2, 4, or 8.
This leaves four possibilities (keeping in mind that (Z/24Z) / ker φ ≅ Im φ and so 24/|ker φ| = |Im φ|):
|φ(g)| = 1 ==> |Im φ| = 1 and thus |ker φ| = 24/1 = 24.
|φ(g)| = 2 ==> |Im φ| = 2 and thus |ker φ| = 24/2 = 12.
|φ(g)| = 4 ==> |Im φ| = 4 and thus |ker φ| = 24/4 = 6.
|φ(g)| = 8 ==> |Im φ| = 8 and thus |ker φ| = 24/8 = 3.
I hope this helps!