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I am trying to prove that if $$ H \subseteq G, $$ where G is a finite group, and $H$ is closed under multiplication, then $H$ is a subgroup of $G$.

We have that $G$ is a finite group. Then, there exist elements $$ a^s = a^r,$$ both in $H$ (because $H$ is closed under multiplication). Then "there must be" that $$ a^{r-s} = e.$$ I don't get this last statement.

I also don't understand why the proof can not be like this next one:

$G$ is a finite group. Then, for every $$ a \in G $$ there exists its inverse. Then, as $H$ is closed under multiplication, $$ aa^{-1}=e .$$ Then $$ e \in H. $$ Associativity is valid in $G$. In particular it is valid for the elements in $H$. Therefore, $H$ is a subroup of $G$.

Thomas Andrews
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Trux
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  • The statement $a^{r-s}=e$ holds only if $G$ (Or $H$) is finite. Also we must have $H\neq\varnothing$ for we'd not have a subgroup, and the closure would hold trivially. That would be: $(\forall a,b\in H)$ (There are none) $ab\in H$. Or If $a,b\in H$ (That's false) then $ab\in H$ (We don't mind if this is true, because the antecedent is false the implication holds trivially.

    There are a lot of hypothesis not in the proof.

     – LeviathanTheEsper Mar 21 '15 at 04:27
  • A Book of Abstract Algebra, Pinter, chapter 5, set D, exercise 5 – dharmatech May 14 '17 at 09:09
  • Similar exercise is also discussed here. – dharmatech May 14 '17 at 19:50

2 Answers2

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The fact that $H$ is closed under multiplication does not guarantee that $a^{-1}\in H$. What can be said is that if $H$ is closed under multiplication, then $a^n\in H$ for every $n\in\Bbb Z^+$.

If $G$ is finite, then closure under multiplication is enough to ensure that $a^{-1}\in H$ for every $a\in H$, since $a^n=e$ for some positive $n$, and so $a^{n-1}=a^{-1}$.

Tim Raczkowski
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Let $G$ be a finite group and H be a subgroup closed under the multiplication. Then $\forall a \in H,\ \exists\ distinct\ i,j\in \mathbb{N}\ s.t. a^i=a^j$

WLOG we can assume that $\ i<j$ which means $j-i \in \mathbb{N}$ and $\ a^{j-i}=e \in H$.

Also you can check that $a^{-1}=a^{j-i-1} \in H$ ( if $j-i-1=0$ then $a=e \in H$).

alemonk
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