Question

for every positive integer $ k $ that divides the order of g, there exists a subgroup $ h $ of $ g $ of order $ k $.

Answer 1

The proof of every positive integer k that divides the order of g, there exists a subgroup h of g of order k is shown below:

What is Subgroup?

A subgroup is a group nested inside another group. The subgroup test is used on subsets of a group to determine if they are subgroups. A proper subgroup is a subgroup that does not contain all of the original group, while a trivial subgroup contains only the identity.

Suppose k ∣ g ⟹n=dk:

If h(g)=⟨g⟩ is cyclic then

clearly K:=|⟨g^d⟩|=k .

Suppose now we have

1≠H≤h(g),

|H|=k and

let r>0 be the minimal natural number s.t.

g^r ∈ H , then

1=(g ^ r)^k = g^(r k)

⟹n ∣ r k ⟹ r k=ns =d ks ⟹ r= ds ⟹ s ∣ d

and from here we have

g^ r=(g^ d)^ s ∈K

Also,  |H|=||K|

we get equality and thus uniqueness of K.

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