Suppose phi is a homomorphism from a group Z36 to a group of order 24.Determine the possible homomorphic images. For each image, determine thecorresponding kernal of phi
here is a hint: the order of the kernel of such a homomorphism has to be a common divisor of 36 and 24. this kernel is a subgroup of Z36. the possible common divisors of 36 and 24 are the possible divisors of 12, namely: 1,2,3,4,6 and 12.
a property of cyclic groups is: if d is a divisor of the order of a cyclic group G, G has EXACTLY one subgroup of order d.
{0}, the unique subgroup of Z36 of order 1.
{0, 18}, the unique subgroup of Z36, of order 2.
{0, 12, 24}, the unique subgroup of order 3.
{0, 9, 18, 27}, the unique subgroup of order 4.
{0, 6, 12, 18, 24, 30} the unique subgroup of order 6.
{0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33}, the unique subgroup of order 12.
where can you go from here?
well, it's better to get away from "element" properties, and think of "group" properties. if φ:Z36→G is a group homomorphism, then ker(φ) is a subgroup of Z36, which has order 36. by lagrange's theorem, the order of ker(φ) has to divide 36. but Z36/ker(φ) is isomorphic to φ(Z36), and (also by lagrange) |φ(Z36)| has to divide |G|.
so if k = |ker(φ)|, we get:
36/k = 1,2,3,4,6,8,12 or 24.
since 36/k is an integer, k = 1,2,3,4,6,9,12,18, or 36.
k = 1 won't work, 36 doesn't divide 24.
k = 2 won't work, 18 doesn't divide 24.
k = 3 would work, 12 divides 24.
k = 4 won't work, 9 doesn't divide 24.
k = 6 would work, 6 divides 24.
k = 9 would work, 4 divides 24.
k = 12 would work, 3 divides 24.
k = 18 would work, 2 divides 24.
k = 36 would work, 1 divides 24.
so the order of the kernel must be 3,6,9,12,18 or 36.
since Z36 is cyclic, φ(Z36) must also be cyclic, with order |φ(1)| in G.
(EDIT: it appears my previous post was wrong, i was thinking of the order of Z36/ker(φ) (or φ(Z36), which is isomorphic to it), which has to divide 36 and 24.
so the possible choices of ker(φ) are:
ker(φ) = {0,12,24} --> φ(Z36) is isomorphic to Z12
ker(φ) = {0,6,12,18,24,30} --> φ(Z36) is isomorphic to Z6
ker(φ) = {0,4,8,12,16,20,24,28,32} -->φ(Z36) is isomorphic to Z4
ker(φ) = {0,3,6,9,12,15,18,21,24,27,30,33} -->φ(Z36) is isomorphic to Z2
ker(φ) = {0,2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34} -->φ(Z36) is trivial.
without knowing more about which group of order 24 we have, you can't really say much more).