Problem statement

Let $G$ be a group and $H \subset G$ a subset. Prove that the following statements are equivalent

(i) $H$ is a subgroup

(ii) $H$ is nonempty and for any $x, y \in H, xy^{-1} \in H$

If $G$ is finite, prove that these statements are equivalent to

(iii) $H$ is nonempty and for any $x,y \in H, xy \in H$

The attempt at a solution

I didn't have problems to show the equivalence of statements (i) and (ii):

$(i) \implies (ii) $ This is by definition, since if $H$ is a subgroup, then is closed under operation, so it satisfies $xz \in H$ for all elements in $H$, in particular, for $z=y^{-1}$.

$(ii) \implies (i)$ is also easy:

To prove the existence of $e \in H$, we can write $e=xx^{-1}$ for any $x \in H$ (we can assure there exists some $x$ since $H$ is non-empty).Now that we've proved the existence of the identity element, by hypothesis, if $x \in H$, then $ex^{-1}=x^{-1} \in H$, so all the elements have inverse in $H$. We prove closure of the operation just by using that $y={y^{-1}}^{-1}$, by hypothesis, $xy=x{y^{-1}}^{-1} \in H$. Associativity follows directly from the fact that $G$ is a group.

Now, $(i),(ii) \implies (iii)$ is immediate, by definition of group, we have that $H$ is closed under the operation, so, for any $x,y \in H, xy \in H$.

I got stuck trying to show that $(iii)$ implies $H$ is a subgroup. If I could prove that for any $x \in H$, $x^{-1} \in H$, then I would be done but I don't know how to show this. I thought to prove it by the absurd: suppose there is an element $x : x^{-1} \notin H$, then how I could conclude that $G$ cannot be finite?

  • 2
    $\begingroup$ Hint: use the fact that the group is finite. What happens if you keep on multiplying $x$ with itself? $\endgroup$
    – Marc
    Mar 24 '14 at 20:23
  • $\begingroup$ Yes, I thought about it, $x^k \neq e$ by my assumption, but how can I assure that there isn't any other $a \in H, G$ such that $x^k=a$? Sorry if my question is too obvious, but I don't see directly that I have infinite distinct elements multiplying $x$ with itself $\endgroup$
    – user100106
    Mar 24 '14 at 20:28
  • 1
    $\begingroup$ if two powers of x were equal, what would that imply? $x^a=x^b$, so $x^{a-b}=e$ contradiction the assumption. $\endgroup$
    – user137500
    Mar 24 '14 at 20:32

If you multiply $x$ with itself repeatedly, then since $H$ is finite you get that two powers must be equal, say $x^a = x^b$ where $a > b$. Then $x^{a-b} = e$ and so $x^{a-b-1} = x^{-1}$ so $x^{-1}$ is in $H$, which as you pointed out is all you need to complete the proof.


Suppose iii) is valid.

If $G$ is finite, for every $x \in H $ we have that $x$ is of finite order , s0 $x^{n} = 1 $ for $n \in \mathbb{N}$.

So $1 = x^{n} \in H$ by hypotheses. Furthermore if $n > 1 $ we have $x^{n-1} = x^{-1} \in H $. This, with the other hypotheses of point iii) proves that $H$ is a subgroup.


Condition (iii) can be relaxed to “$H$ is finite”. Let $y\in H$ (which exists because $H$ is non empty); then the map $$ f_y\colon H\to H,\quad x\mapsto xy $$ is well defined by hypothesis. This map is injective, for $$ f_y(x_1)=f_y(x_2)\implies x_1y=x_2y\implies x_1yy^{-1}=x_2yy^{-1} \implies x_1=x_2 $$ since an inverse for $y$ exists in $G$. Therefore $f_y$ is also surjective, since $H$ is finite and so there is $x_0\in H$ with $f_y(x_0)=y$, which means $x_0=1\in H$. Next, there is $x'\in H$ such that $f_y(x')=1$, which means $x'=y^{-1}\in H$.

Since $y$ is an arbitrary element of $H$, we have proved that $H$ contains the inverse of each of its element.


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