Let G is cyclic group and |G|=48. How do you find all of subgroup of G ?

1 Answer
Feb 23, 2018

The subgroups are all cyclic, with orders dividing 48

Explanation:

All subgroups of a cyclic group are themselves cyclic, with orders which are divisors of the order of the group.

To see why, suppose G= < a > is cyclic with order N and H sube G is a subgroup.

If a^m in H and a^n in H, then so is a^(pm+qn) for any integers p, q.

So a^k in H where k = GCF(m, n) and both a^m and a^n are in < a^k >.

In particular, if a^k in H with GCF(k, N) = 1 then H = < a > = G.

Also not that if mn = N then < a^m > is a subgroup of G with order n.

We can deduce:

  • H has no more than 1 generator.
  • The order of H is a factor of N.

In our example N = 48 and the subgroups are isomorphic to:

C_1, C_2, C_3, C_4, C_6, C_8, C_12, C_16, C_24, C_48

being:

< >, < a^24 >, < a^16 >, < a^12 >, < a^8 >, < a^6 >, < a^4 >, < a^3 >, < a^2 >, < a >