Prove that subgroup of cyclic group is cyclic.

Understand the Problem

The question is asking for a mathematical proof that any subgroup of a cyclic group is also cyclic. This involves understanding the properties of cyclic groups and subgroups.

Answer

Any subgroup of a cyclic group is cyclic.

Steps to Solve

Define a Cyclic Group

A cyclic group is defined as a group that can be generated by a single element. Let's denote a cyclic group by $G$, which is generated by an element $g$. This can be written as: $$ G = \langle g \rangle = { g^n \mid n \in \mathbb{Z} } $$

Consider a Subgroup of G

Let $H$ be a subgroup of the cyclic group $G$. Since $H$ is a subgroup, it must also consist of elements of $G$.

Find an Element in H

Since $G$ is generated by $g$, let's choose an element $h \in H$. Since $G$ is infinite or finite, the element $h$ must be of the form $g^k$ for some integer $k$.

Show H is Generated by h

Now, let’s show that the subgroup $H$ can be generated by the element $h$. We can write: $$ H = { g^{nk} \mid n \in \mathbb{Z} } $$

Conclude that H is Cyclic

Since all elements in $H$ can be expressed in the form of $g^{nk}$, this means that $H$ is generated by the single element $h$, which shows that $H$ is cyclic. Therefore, any subgroup of a cyclic group is cyclic.

Any subgroup of a cyclic group is also cyclic.

More Information

Cyclic groups are important in group theory, a branch of mathematics that studies algebraic structures called groups. They are fundamental in many areas including number theory, algebra, and geometry.

Tips

Forgetting that the chosen element $h$ must belong to the subgroup $H$. It’s essential to ensure that $h$ is actually an element of $H$ to prove that $H$ is cyclic properly.
Assuming that any subgroup can contain elements outside of the set generated by a single group element.

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