Prove that the subgroup generated by $S$ is equivalent to the set of all integral powers of elements of $S$

Question

(Exercise C.1 Lee’s Introduction to Topological Manifolds) Suppose $G$ is a group and $S\subseteq G$. Prove that the subgroup generated by $S$ is equivalent to the set of all finite products of integral powers of elements of $S$.

My attempt: Let $H$ be defined as $$H = \left\{\prod_{i=1}^n a_i^{k_i} \ \bigl\vert \ \{a_i\}_{i\in\Bbb N}\in S, \{k_i\}_{i\in\Bbb N}\in\Bbb Z\right\}.$$ Obviously, we have $a_i^{k_i}\in \langle S\rangle, \ \forall i\in \Bbb N_0$ since any group is closed under its operation and thus, $H\subseteq \langle S\rangle$.

Now, we must show that $\langle S\rangle \subseteq H$. It suffices to prove that $S\subseteq H \le G$, since $\langle S\rangle$ is the smallest subgroup containing $S$. Note $H$ must contain the identity since in the case that $i\equiv 0$, $$\left\{\prod_{i\in\varnothing}a_i^{k_i} \ \bigl\vert \ \{a_i\}_{i\in\Bbb N}\in S, \{k_i\}_{i\in\Bbb N}\in \Bbb Z\right\} = \{1\},$$ thus $H\le G$. Too, $S\subseteq H$ [insert explanation here]. $\quad \square$

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Why should $S\subseteq H$ — how can I complete my proof?

Also, I saw this recommended question and noticed that the answers gave a similar proof to my attempt above. The only difference is, they restrict their attention to powers in the set $\{-1,1\}$, while my exercise concerns itself with all integral powers. So, I wonder, is Exercise C.1 a corollary to wmnorth’s problem? Clearly, mine implies his, but is the converse true?

Thanks in advance.

Answer 1

The equivalence of your problem and wmnorth's is easy: if you can write an element of $\langle S\rangle$ as products of powers of elements of $S$, then each power $s^k$ can be expanded like $s\cdot s\cdots s$ or $s^{-1}\cdot s^{-1}\cdots s^{-1}$ as a product of powers with exponents $\pm1$.

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Some quibbles about your post:

1. In your definition of $H$, your are missing the condition that each $a_i\in S$.
2. $a_i^k$ is not in $S$. It is in $\langle S \rangle$. $S$ is not a subgroup, there is no reason for it to be closed under the group operation.
3. $H\subseteq S$ is not true, for the same reason as 2.
4. $S\subseteq H$ is trivially true.

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To show $H=\langle{S\rangle}$, you need to show two things:

1. $H$ is a subgroup.
2. Any subgroup $K$ of $G$ containing $S$ contains $H$.

(1) To find the product of two elements in $H$, just concatenate the lists $\{a_i\}$ and $\{k_i\}$. For inverses, negate all of the $k_i$ and reverse the order of the elements. As you noted, the empty product shows $1\in H$.

(2) If $K$ is a subgroup of $G$ containing $S$, then the closure of $K$ under multiplication and inversion implies $K$ contains all products of elements in $S$ in their inverses, which includes all of $H$.

Answer 2

$\langle S \rangle \subseteq H$ because $H$ is a subgroup. Note that if you multiply two elements of $H$, the results is still an element of $H$ (i.e., still a product of integral powers of elements of $S$..just juxtapose). The same is true of the inverse of any element of $H$...just write in reverse order and negate all the exponents.