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

Question

3136 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-23T07:26:10

The subgroups are all cyclic, with orders dividing #48#

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:

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 >#