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Let $G$ be a group, and $H$ a subset of $G$.

If $H$ is a subgroup of $G$, then $H^2 = H$. I'm interested in what holds if $H^2 = H$. It's easy to verify that $H$ is a semigroup. If $H$ is finite, $H$ is also a monoid:

Assume $e \notin H$. Fix any $h \in H$. Then $h \notin hH$. Since $hH \subseteq H^2 = H$, we have $|hH| < |H|$, a contradiction. Thus $e \in H$.

If $H$ is infinite, then $H$ needn't be a monoid: $(\mathbb{Q}, +)$ is a group, and $(\mathbb{Q}^+, +)$ isn't a monoid, even though $\mathbb{Q}^+ + \mathbb{Q}^+ = \mathbb{Q}^+$.

But what if $H$ is finite; is $H$ perhaps a group?

Thank you!

2 Answers 2

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If $H$ is finite, then $H$ is a group.

Note that if $h \in H$, then $\{h,h^2,h^3,\ldots\} \subset H$, which implies that $h$ must have finite order (say $n$). In particular, $h^{-1} = h^{n-1} \in H$, so $H$ is closed under taking inverses. Since also $e \in H$ and $H$ is closed under multiplication, $H$ is a group.

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    This is misleadingly phrased. $h$ necessarily has finite order since it has finite order in $G$. You probably mean to say "this implies that the orbit of $h$ under the action is entirely contained within $H$" – 2017-01-04
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    $G$ is not assumed to be finite, so I do not think anything can be said a priori about the order of $h$ in $G$ – 2017-01-04
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    ah, reading failure on my part – 2017-01-04
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Here's a slight variation to your proof. Fix $h\in H$. Consider the injection $f: hH\to H$. Since both sets are finite and $|hH| = |H|$, $f$ is a bijection. Hence $e\in H$ has a preimage under $f$, which means there exists a $h'\in H$ such that $hh' = e$, so $H$ is closed under taking inverses.