The Subgroup Generated By An Element Is Actually A Subgroup Tyler

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Problem 58 (The Subgroup Generated By An Element Is Actually A Subgroup)

Let $G$ be a group with $a\in G$. Show that $\langle a\rangle$, using the definition above, is a subgroup of $G$.

Solution

In order to show that $\langle a \rangle \leq G$ we must show the following: $\langle a \rangle \subseteq G$ where $\langle a \rangle \neq \emptyset$, if $a, b \in \langle a \rangle$, then $ab \in \langle a \rangle$, and if $a \in \langle a \rangle$, then $a^{-1} \in \langle a \rangle$.

By the definition of a subgroup of $G$ generated by an element of $G$, we know that if $a \in G$, then $a \in \langle a \rangle$, since $a^1 = a \in \langle a \rangle$. Therefore, $\langle a \rangle \neq \emptyset$. Then let $b \in \langle a \rangle$. This means that $b = a^k$ for some $k \in \mathbb{Z}$. Since $G$ is a group, and since $a \in G$ it follows that $a^k \in G$. Thus $\langle a \rangle \subseteq G$.

Next consider $a^i, a^j \in \langle a \rangle$ for some $i, j \in \mathbb{Z}$. We compute $$ \begin{align} a^i a^j &= \underbrace{aa \cdots a}_{i} \underbrace{aa \cdots a}_{j} \\ &= \underbrace{aa \cdots a}_{i + j} \\ &= a^{i + j}. \end{align} $$ Since $i + j \in \mathbb{Z}$ it follows by the definition of a subgroup generated by $a$ that $a^{i + j} \in \langle a \rangle$; therefore, $\langle a \rangle$ is closed.

Finally, consider $c \in \langle a \rangle$. This mean that $c = a^m$ for some $m \in \mathbb{Z}$. Since $c \in G$ as well, we know that $c^{-1} = a^{-m} \in G$. And clearly since $a^{-m} \in \langle a \rangle$, we have that $c^{-1} \in \langle a \rangle$.

Q.E.D.

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