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I wrote this question from Hungerford's introduction abstract p.33 algebra

Theorem 2.8: If $G$ is a group and $X$ is a nonempty subset of $G$ then the subgroup $\langle X\rangle$ consists of all finite products $ a_{1}^{n_{1}}a_{2}^{n_{2}}\cdots a_{t}^{n_{t}}|(a_{i} \in X;n_{i} \in \mathbb{Z})$. In particular for every $ a\in G$, $$\langle a\rangle=\{a^{n}|n \in \mathbb{Z}\}$$

I can't show this.I know $\langle X\rangle=\bigcap_{i \in I}H_i,$ where $\{H_{i}|i\in I\}$ is family of all subgroups of $G$ which contain $X$. Therefore we can easily see $$\{ a_{1}^{n_{1}}a_{2}^{n_{2}}\cdots a_{t}^{n_{t}}|(a_{i} \in X;n_{i} \in \mathbb{Z})\}\subseteq \langle X\rangle$$

  1. How can we show converse direction?

  2. What is the meaning of this theorem? For example: $\mathbb Z\setminus \{0\}$ is a subset of $\mathbb Q\setminus \{0\}$ and $\mathbb Q\setminus \{0\}$ is a subset of $\mathbb R\setminus \{0\}$. How can we find set $\bigcap_{i \in I}H_i $ for those examples?

    If we think group as a additive group, Then how do we determine the above concept?

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This theorem is the non-commutative, group analogue of the linear algebra fact that the subspace generated by a set $X$ is the set of finite linear combinations of elements of $X$. The proof follows the same lines: prove that the set of finite linear combinations of elements of $X$ is a subspace that contains $X$, and that every subspace containing $X$ contains all finite linear combinations of elements of $X$.

What you're missing is the first part: that the set of all finite products of elements of $X \cup X^{-1}$ is a group that contains $X$.

Perhaps it'll easier to state all this as follows:

Let $G$ is a group and $X$ is a nonempty subset of $G$. Then the following are equivalent for a subgroup $H$ of $G$:

  • $H$ is the smallest subgroup of $G$ containing $X$, in the sense that if $H'$ is a subgroup of $G$ containing $X$ then $H \subseteq H'$.
  • $H$ is the intersection of all subgroups of $G$ containing $X$.
  • $H$ is the set of all finite products of elements of $X \cup X^{-1}$.
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  • I know proof concept.How can we say H is the smallest subgroup of G containing X ? – 1ENİGMA1 Dec 14 '15 at 11:43
  • Ok.it is easy ${H_{i}|i\in I}$ all family of subgroup G which containing X.So if any subgroup K containing X must be $H_{i_{0}}$.Hence $\bigcap_{i \in I}H_i \subset H_{i_{0}}$.So $\bigcap_{i \in I}H_i$ is the smallest subgroup containing is X.we showed ${ a_{1}^{n_{1}}a_{2}^{n_{2}}\cdots a_{t}^{n_{t}}|(a_{i} \in X;n_{i} \in \mathbb{Z})}\subseteq \langle X\rangle=\bigcap_{i \in I}H_i $,but $\bigcap_{i \in I}H_i $ the smallest subgrouf of G and so $\bigcap_{i \in I}H_i = \langle X\rangle={ a_{1}^{n_{1}}a_{2}^{n_{2}}\cdots a_{t}^{n_{t}}|(a_{i} \in X;n_{i} \in \mathbb{Z})}$ – 1ENİGMA1 Dec 14 '15 at 11:55
  • This is really reasonable for example additive group ($\mathbb{Q},+$) if infinite summation is meaningful then, $\frac{1}{0!}+ \frac{1}{1!}+\frac{1}{2!}+...=e \notin \mathbb{Q}$ – 1ENİGMA1 Dec 14 '15 at 12:15
  • How can we use this theorem. $2 \mathbb{Z}$ is a nonempty subset of $ \mathbb{Z}$ ( $ \mathbb{Z},+$)is a group .So we can say as a theorem $ \langle 2Z \rangle={2k_{1}.n_{1}...2k_{t}n_{t}| 2k_{i} \in 2 \mathbb{Z},n_{i} \in \mathbb{Z}}={2k:k \in \mathbb{Z}}$ is it true ? – 1ENİGMA1 Dec 14 '15 at 12:49