# Let $G$ be a group such that there exists $a\in G$ such that $H = G \setminus\{a\}$ is a subgroup of $G$. Show that $|G|=2$.

Let $G$ be a group such that there exists $a\in G$ such that $H = G \setminus\{a\}$ is a subgroup of $G$. Show that $|G|=2$.

This problem was given as my homework in Abstract Algebra. But I don't seem to find a way to prove this without using Lagrange's Theorem (order of subgroup must divide the order of the group that the subgroup is in) because I haven't learned that yet in class.

I also don't know if I should assume $G$ is finite or cyclic in order to show $|G|=2.$ Should I first show that the group $G$ is finite first?

• You can follow some of the proof of Lagrange's theorem. Show that, if $h \in H$, then $ha \notin H$, hence $ha = a$. – Theo Bendit Feb 24 '18 at 23:14
• @TheoBendit But $ha\notin H$ because $a\notin H$ as the definition of $H=G\setminus \{a\}$ for some $a\in G$. I am curious why are we allowed to choose that specific $a\in G$ rather than other elements that could be in $H$, like $c,d,... \in H$ taking the possibility that $H$ could have more elements than $a$ and $b$? – user3000482 Feb 27 '18 at 1:09
• The $a$ element I'm taking is the one in the question: the one and only element of $G$ that is not in $H$. Certainly, if I pick $b$, some element of $G$ other than $a$, then $b \in H$, so $bh \in H$ by closure under the group operation. I picked on $a$ because it lies outside $H$, and so the coset $Ha$ is not equal to $H$. I don't know if you've covered cosets yet, but you can still show rather directly that $ha \notin H$. – Theo Bendit Feb 27 '18 at 1:14
• @TheoBendit Could you look at the last comment I added to the answer of this question? I am curious why $ab^{-1}=a$ after knowing that $ab^{-1}\notin H$. Why can't $ab^{-1}=c$ for some $c\in H$? Because we don't know yet how many elements $H$ can have. – user3000482 Feb 27 '18 at 1:33
You can approach it by thinking about the consequences of putting a hole in the multiplication table: assume $H = G \setminus \{a\}$ is a subgroup of $G$ and let $b \in H$. I claim that $ab^{-1} \not\in H$, for if $ab^{-1} \in H$, then as $b \in H$ and $H$ is a subgroup, then $a = (ab^{-1})b \in H$, contradicting the definition of $H$ as $G \setminus \{a\}$. So we must have $ab^{-1} = a$, implying that $b = 1$. So $H = \{1\}$ and $G = \{1, a\}$.
• How can you let $ab^{-1} \in H$ when we know that $a\notin H$? Aren't you using the $1$-step subgroup test? Could you elaborate more on the step where you make the contradiction of having $a=(ab^{-1})b\in H$? – user3000482 Feb 25 '18 at 2:10
• But how do you know that $ab^{-1}=a$? Why is it not possible for $ab^{-1}=c$ for some $c\in H?$ It seems to me that you are proving as if you already know that $G$ only has two elements. Since we don't know yet $|G|=2$, shouldn't we also account for other elements that could be in $G$ other than $a,b$? Can you please explain your proof so I can follow? – user3000482 Feb 25 '18 at 23:17