1
$\begingroup$

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?

Please help me on how I should approach this problem.

$\endgroup$
  • 2
    $\begingroup$ You can follow some of the proof of Lagrange's theorem. Show that, if $h \in H$, then $ha \notin H$, hence $ha = a$. $\endgroup$ – Theo Bendit Feb 24 '18 at 23:14
  • $\begingroup$ @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 $b$? $\endgroup$ – user3000482 Feb 27 '18 at 1:09
  • $\begingroup$ 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 $ha \notin H$. $\endgroup$ – Theo Bendit Feb 27 '18 at 1:14
  • $\begingroup$ @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. $\endgroup$ – user3000482 Feb 27 '18 at 1:33
2
$\begingroup$

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 $G = \{1, a\}$.

$\endgroup$
  • $\begingroup$ 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 $a=(ab^{-1})b\in H$? $\endgroup$ – user3000482 Feb 25 '18 at 2:10
  • $\begingroup$ 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? $\endgroup$ – user3000482 Feb 25 '18 at 23:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.