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In the book Theory of Finite Groups by Kurzweil and Stellmacher we find on page 5 the following statement and proof:

A nonempty finite subset $U$ of $G$ is a subgroup if for all $x,y \in U$ also $xy$ is in $U$.

Proof. For $x \in U$ the mapping $\varphi : u \mapsto ux$ from $U$ to $U$ is injective and thus also surjective since $U$ is finite. If follows that $1 = x^{\varphi^{-1}} \in U$ and $x^{-1} = 1^{\varphi^{-1}} \in U$. $\square$

I have no doubt about the statement, but I guess there is a subtle inaccuracy in its proof. How do we know that the inverse image of $x$ by $\varphi^{-1}$ is $1$. Certainly as $\varphi$ is bijective on $U$ we have $x^{\varphi^{-1}} \in U$.

So why $x^{\varphi^{-1}} = 1$?

Remark: For me a "correct" proof would go like this: Consider $\varphi : G \to G$ with $g \mapsto gx$, then this map is injective and as $\varphi(1) = x$ we have $x^{\varphi^{-1}} = 1$. Furthermore the restriction to $U$ is also injective and therefore surjective on $U$ as its image lies always in $U$, and this gives $1 \in U$ as $1$ must be its unique inverse image. This proof considers the map on the whole of $G$ first, but the above mentioned proof considers a map $U \to U$ from first principles, and I see no way to derive from that $x^{\varphi^{-1}} = 1$; for example imagine that an extension on $G$ is not injective, then an inverse image of $x$ in this extension mapping might be $1$, and another one might be the element in $U$ different from $1$.

4 Answers 4

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The proof is valid:

$1=(1^{\varphi^{-1}})^\varphi=1^{\varphi^{-1}}x$ so $1^{\varphi^{-1}}=x^{-1}$

Alternatively, you can check $\varphi^{-1}:u\mapsto ux^{-1}$ so $1^{\varphi^{-1}}=x^{-1}$

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    But how to evaluate $1^{\varphi^{-1}}$ when you do not know that $1 \in U$ in the first place? I mean $\varphi^{-1}$ is just defined on elements of $U$... – 2017-01-14
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    For your alternative way, as we do not know that $x^{-1} \in U$ yet, we cannot assume that the map $u \mapsto ux^{-1}$ is from $U$ to $U$, so I have my doubts about that... – 2017-01-14
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    the proof given includes $1=x^{\varphi^{-1}}\in U$ – 2017-01-14
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    And specifically about that is my question! – 2017-01-14
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    We know $\varphi$ is bijective, $\varphi^{-1}$ exists. For $v\in U$ surjectivity implies $v=ux$ for some $u\in U$ so $v^{\varphi^{-1}}=(ux)^{\varphi^{-1}}=u=vx^{-1}$. You need not assume $x^{-1}\in U$ – 2017-01-14
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    Okay, thats clear. – 2017-01-14
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We know that $e:=x^{\phi^{-1}}\in U$ is an element of $U$ with the property $ex=e^\phi=x$. As $G$ is a group, we conclude that $e=1$.

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Well you know that by definition $x^{\varphi^{-1}}x=(x^{\varphi^{-1}})^\varphi=x$ in the group $G$. This clearly implies $x^{\varphi^{-1}}=1$. I wouldn't call this an inaccuracy, it's just not doing all the small details, which is extremely frequent in math books...

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Always remember that you're working in a group $G$, so whenever necessary, we can bounce our computations up to the group itself.

Sketch:

  1. $\varphi$ is injective because if $ux=vx$, then applying $x^{-1}$ from $G$ gives you $u=v$.

  2. An injective map from a finite set to itself must be surjective because there are only finitely many elements (and you run out of elements otherwise). Therefore, $\varphi$ is surjective.

  3. Since $\varphi$ is surjective, there is some $w$ in $U$ so that $\varphi(w)=x$ (this happens because $x\in U$ and $\varphi$ is surjective).

  4. Observe that this means $wx=x$, so applying $x^{-1}$ from $G$ gives you $w=1$. Therefore, $U$ must contain the identity of $G$.

  5. Moreover, we see that since $\varphi(1)=x$, $\varphi^{-1}(x)=w=1$.