The book I'm using gives the following test for checking if a subset of a group is a subgroup:

Let $G$ be a group and let H be a nonempty subset of $G$. If $ab$ is in $H$ whenever $a$ and $b$ are in $H$ ($H$ is closed under the operation), and $a^{-1}$ is in $H$ whenever $a$ is in $H$ ($H$ is closed under taking inverses), then $H$ is a subgroup of $G$.

It uses this to prove the following statement:

Let $H$ be a nonempty finite subset of a group $G$. If $H$ is closed under the operation of $G$, then $H$ is a subgroup of $G$.

It provides the following proof:

In view of the above theorem, we need only prove that $a^{-1} \in H$ whenever $a \in H$. If $a=e$, then $a^{-1}=a$ and we are done. If $a \neq e$, consider the sequence $a$, $a^2$,.... By closure, all of these elements belong to $H$. Since H is finite, not all of these elements are distinct. Say $a^i=a^j$ and $i \geq j$. Then, $a^{i-j}=e$; and since $a \neq e$, $i - j > 1$. Thus, $aa^{i-j-1}=a^{i-j}= e$ and, therefore, $a^{i-j-1}=a^{-1}$. But $i-j-1>1$ implies $a^{i-j-1}\in H$ and we are done.

I understand the proof, but before reading it, I tried proving this myself. An my proof seems very different:

  1. First check $e \in H$ -- $H$ is non-empty. So there exists $a \in H$. H is closed, so it is true that for $x,y\in H$ (not necessarily different from $a$) that $x*y \in H$. In particular consider $a*y=a$. Then $y=e$. So $e \in H$
  2. Now check $ab \in H$ -- H is closed so $ab\in H$ whenever $a \in H$ and $b \in H$.
  3. Finally check $a^{-1} \in H$ -- By 1, $a$ and $e$ are in $H$. Consider $a*x=e$. Thus $x = a^{-1}$. We know $H$ is closed. So $a^{-1} \in H$.

My proof didn't use the finiteness of $H$, so there's a flaw somewhere - but I can't spot it. Where did I use finiteness?

  • 1
    $\begingroup$ What proof do you have that $a * x = e$ has a solution $x$ in $H$? $\endgroup$ Nov 10, 2017 at 14:03

1 Answer 1


Being closed under products means that "for all $x,y\in H$, $xy\in H$"; but in 1 and 3 what you use is rather "for all $x\in H$, if $xy\in H$ then $y\in H$". This is not equivalent; for example, the subset $\Bbb N$ of $\Bbb Z$ is closed under addition, but it does not have the property you use, because for example $3+(-2)=1\in \Bbb N$ and $3\in \Bbb N$ but $-2\notin \Bbb N$.

In fact, the property you want to use is equivalent to saying that "for all $x,y\in H$, $x^{-1}y\in H$", which is a classical criterion to determine if a non-empty subset is a subgroup.


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