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

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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:

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

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