5
$\begingroup$

Let $G$ be a finite group, and let $S$ be a nonempty subset of $G$. Suppose $S$ is closed with respect to multiplication. Prove that $S$ is a subgroup of $G$. (Hint: It remains to prove that $S$ contains $e$ and is closed with respect to inverses. Let $S = \{a_{1},...,a_{n} \}.$ If $a_{i} \in S$, consider the distinct elements $a_{i}a_{1}, a_{i}a_{2},...a_{i}a_{n}$.)

Preliminary questions: Can I assume that every element in $S$ unique? If each element is unique, then does that imply the products $a_{i}a_{1}, a_{i}a_{2},...a_{i}a_{n}$ are distinct? If I cannot assume that each element is not unique, then why should I just consider the distinct products?

(i). Show that $S$ contains $e$: Because we know $S$ is closed under multiplication, we know that each element in $S$ can be written as a product of elements in $S$. The book claims that I can write $a_{1} = a_{1}a_{k}$ for some $k$. I don't really understand why I can make such a claim. Does the following explain why? If $a_{1} \in S$ then will one of the following products $a_{1}a_{1}, a_{1}a_{2}, ... , a_{1}a_{n}$ be equal to one element in $S$ and eventually will every product be paired with each element in $S$? How do we know $a_{1} \neq a_{3}a_{2}$?

Because if I assume $a_{1} = a_{1}a_{k}$, then $a_{1}^{-1}a_{1} = a_{1}^{-1}a_{1}a_{k}$ shows that $e = a_{k}$. And I can still make use of $a_{1}^{-1}$ because if $a_{1} \in S$, then $a_{1} \in G$ and $a_{1}^{-1} \in G$ since $G$ is a group. Is this reasoning correct?

(ii). Show that every element $a_{p} \in S$ has an inverse $a_{p}^{-1} \in S$: Let $a_{p} \in S$, then we can write $a_{p} = a_{1}a_{q}$ for some $q$. We can rearrange and get $a_{q} = a_{1}^{-1}a_{p} = (a_{p}^{-1}a_{1})^{-1}$. At this point I've shown that $(a_{p}^{-1}a_{1})^{-1} \in S$, but now I have been stuck for a while. Could I get a hint?

Thanks in advance.


Update

(i). Show that $S$ contains $e$: Using the hint, ``If $a_{i} \in S$, consider the distinct elements $a_{i}a_{1}, a_{i}a_{2},...a_{i}a_{n}$," if $a_{1} \in S$, then we consider $a_{1}a_{1}, a_{1}a_{2},...a_{1}a_{n}$. Then we can write the following: \begin{align*} a_{1} &= a_{1}a_{k} \newline a_{1}^{-1}a_{1} &= a_{1}^{-1}a_{1}a_{k} \newline e &= ea_{k} \newline e &= a_{k} \end{align*}

(ii). Show that every element $a_{p} \in S$ has an inverse $a_{p}^{-1} \in S$: To show that $e \in S$, we can write $a_{1}a_{p} = e$ for $a_{1}, a_{p} \in S$ \begin{align*} a_{1}a_{p} &= e\newline a_{1}a_{p}a_{p}^{-1} &= ea_{p}^{-1}\newline a_{1} &= a_{p}^{-1} \end{align*}

  • 3
    The notation $S = \{a_1, \ldots, a_n\}$ implies the $a_i$ are distinct elements. Just an example of how to proceed : $a_1 a_2$ and $a_1 a_3$ are distinct because if the were equal, then you'd get $a_2 = a_3$ (by multiplying by $a_1^{-1}$, which exists whether it's in $S$ or not). With the same kind of reasoning the products $a_i a_1, \ldots, a_i a_n$ are distinct.2011-08-18
  • 3
    @Joel: I would disagree that the notation implies that the elements are necessarily distinct, but there is no harm in assuming they *are* distinct, since, e.g., $\{1,2,2\} = \{1,2\}$.2011-08-18
  • 0
    @Arturo : True :)2011-08-20
  • 0
    *A Book of Abstract Algebra*, Pinter, chapter 5, set D, exercise 52017-05-14
  • 0
    Similar exercise is also discussed [here](https://math.stackexchange.com/questions/1199227/prove-that-h-is-a-subgroup-of-g).2017-05-14

3 Answers 3