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Here are my thoughts so far: in order for it to be a subgroup it has to have the identity and it has to follow the operation. Lets say we take away an element. Then two other elements in the group that when the operation is performed on them make up that element therefore it is not a group.

My other idea is that every element in the group has to have an inverse in the group. Therefore if we take away an element then one elements inverse goes away making it not a group or you would have to take away the identity element to avoid that but then the identity element is gone therefore it is not a group.

I think the second one is closer to a proof but I'm not sure how to put it into a proof form, or is this okay?

Shaun
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Allie
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  • It really makes sense to have a look at Lagrange's theorem before doing this exercise (even if it was not yet proved in class). It makes it so clear. – Dietrich Burde Sep 23 '20 at 15:31

4 Answers4

4

Hint:

Suppose $G$ has a subgroup of order $n-1$. Call it $H$. Then there is a unique element $y \in G$ with $y \not\in H$.

Pick some non-identity element $x \in H$.

Is $xy \in H$?

Brian Moehring
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Hint: Lagrange's theorem. That's, the order of the subgroup must divide $n$.

In fact, it's easy to see that $\rm{gcd}(n,n-1)=1$.

1

The second one doesn't work too well, since it doesn't work if all elements are their own inverses. For the first one, all you need to do is, for any $a\in G$, find some $b,c\in G$ that are not equal to $a$ so that $bc=a$. Pick $b$ arbitrarily; can you find some $c$ that does what you need it to?

Carl Schildkraut
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By Lagrange's Theorem, if such a subgroup existed, then $n-1\mid n$. This does not happen for $n>2$. (See here.)

Shaun
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