6. Let $$ G $$ be an abelian group of order $$ m $$. If $$ n $$ divides $$ m $$, prove that $$ G $$ has a subgroup of order $$ n $$.

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We start with the Fundamental Theorem of Finite Abelian Groups, which tells us that every finite abelian group $$ G $$ of order $$ m $$ is isomorphic to a direct product of cyclic groups of prime power order. In other words, there exist prime numbers $$ p_1, p_2, \ldots, p_k $$ and positive integers $$ a_1, a_2, \ldots, a_k $$ such that

$$
G \cong \mathbb{Z}_{p_1^{a_1}} \times \mathbb{Z}_{p_2^{a_2}} \times \cdots \times \mathbb{Z}_{p_k^{a_k}}
$$

Let $$ n $$ be a divisor of $$ m $$. Since $$ m = |G| $$, we may write

$$
m = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} \quad \text{and} \quad n = p_1^{b_1} p_2^{b_2} \cdots p_k^{b_k}
$$

with $$ 0 \le b_i \le a_i $$ for each $$ i $$.

The idea is to construct, for each prime $$ p_i $$, a subgroup of $$\mathbb{Z}_{p_i^{a_i}}$$ of order $$ p_i^{b_i} $$ and then take the direct product of these subgroups. We note that every cyclic group $$\mathbb{Z}_{p_i^{a_i}}$$ has a unique subgroup of order $$ p_i^{b_i} $$ (this is a classical fact about cyclic groups).

For each $$ i $$, let $$ H_i $$ be the subgroup of $$\mathbb{Z}_{p_i^{a_i}}$$ of order $$ p_i^{b_i} $$ (with the convention that if $$ b_i = 0 $$, we take $$ H_i $$ to be the trivial subgroup).

Now consider the subgroup

$$
H = H_1 \times H_2 \times \cdots \times H_k
$$

Since the order of a direct product is the product of the orders of the factors, we have

$$
|H| = |H_1| \cdot |H_2| \cdots |H_k| = p_1^{b_1} p_2^{b_2} \cdots p_k^{b_k} = n.
$$

Thus, $$ H $$ is a subgroup of $$ G $$ with order $$ n $$, completing the proof.
To prove that an abelian group $$ G $$ of order $$ m $$ has a subgroup of order $$ n $$ when $$ n $$ divides $$ m $$, follow these steps: 1. **Express $$ G $$ as a Direct Product of Cyclic Groups**: By the Fundamental Theorem of Finite Abelian Groups, $$ G $$ can be written as $$ \mathbb{Z}_{p_1^{a_1}} \times \mathbb{Z}_{p_2^{a_2}} \times \cdots \times \mathbb{Z}_{p_k^{a_k}} $$, where $$ p_i $$ are prime numbers and $$ a_i $$ are positive integers. 2. **Factor $$ n $$ into Prime Powers**: Since $$ n $$ divides $$ m $$, we can write $$ n = p_1^{b_1} p_2^{b_2} \cdots p_k^{b_k} $$, with $$ 0 \leq b_i \leq a_i $$ for each $$ i $$. 3. **Construct Subgroups for Each Cyclic Component**: For each $$ i $$, find a subgroup $$ H_i $$ of $$ \mathbb{Z}_{p_i^{a_i}} $$ of order $$ p_i^{b_i} $$. If $$ b_i = 0 $$, $$ H_i $$ is the trivial subgroup. 4. **Form the Direct Product of Subgroups**: Let $$ H = H_1 \times H_2 \times \cdots \times H_k $$. The order of $$ H $$ is $$ |H| = p_1^{b_1} p_2^{b_2} \cdots p_k^{b_k} = n $$. 5. **Conclusion**: $$ H $$ is a subgroup of $$ G $$ with order $$ n $$, proving that $$ G $$ has a subgroup of order $$ n $$. Thus, an abelian group of order $$ m $$ that is divisible by $$ n $$ contains a subgroup of order $$ n $$.

Let be an abelian group of order . If divides , prove that has a
subgroup of order .

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