# Subgroup of order $n-1$ of a group of order $n$

Here is question 2.1.5 from Dummit and Foote : Prove that $G$ cannot have a subgroup $H$ with $|H| = n-1$, where $n = |G| > 2$.

How can one show this without using Lagrange's theorem (which is in chapter 3 of Dummit).

Thank you

• If I am not mistaken, Lagrange's theorem part of exercise 19 in section 1.7. So you do have it available to you. Commented Oct 10, 2016 at 14:25

We know that any subgroup must have the identity element in it. We also know that every subgroup must contain inverses for all of its elements. Suppose that $H$ is a subgroup of order $n-1$. Let $x$ designate the one element of $G$ not in $H$. Then $x$ must be its own inverse, as if $x^{-1} \not= x$ we have that $x^{-1}\in H$, yet $x$ is not in $H$ (which is a contradiction).

Now take any non-identity $y\in H$. Then if $xy$ is in $H$, then this implies that $x$ is in $H$ since we can multiply by $y^{-1}$. The only way that $xy$ is not in $H$ is if $y=1$. But we assumed otherwise.

We then reach the contradiction.

Take a look at $gH$, where $g \in G$, $g \notin H$. Then I claim $gH \cap H = \varnothing$; for if $h \in gH \cap H$, then we have $h \in H$ and also $h \in gH$, whence $h = gk$ for $k \in H$. Then $g = hk^{-1} \in H$, a contradiction. So $gH \cap H = \varnothing$; but then since $gH$ and $H$ each have $n - 1$ elements, $H \cup gH$ has $2(n - 1) = 2n - 2 > n$ elements if $n > 2$. This contradicts $\mid G \mid = n$. So such a subgroup $H$ with $\mid H \mid = n - 1$ cannot exist. QED

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

• It is worth observing that the notion that $H \cap gH = \varnothing$ is going in the direction of the coset partition of $G$, which is the key idea in Lagrange's theorem. Commented Dec 3, 2013 at 20:19

Let $g$ be an element of $G$ that is not in $H$. Let $h$ be any nonidentity element of $H$. Now show that $gh$ is not in $H$.

• I'm not sure to understand... $H$ has to be closed under the group operation. Since $g$ is not in $H$, why can we conclude from the fact that $gh$ is not in $H$ ?
– M.G
Commented Dec 3, 2013 at 20:08
• Because if $h \in H, h^{-1} \in H$, so if $gh \in H, ghh^{-1} \in H$. This is a leadin to Lagrange's theorem. Commented Dec 3, 2013 at 20:10
• @M.G. Suppose $gh\in H$. Then $g=(gh)h^{-1}\in H$, a contradiction. Just as Ross says.
– Doc
Commented Dec 3, 2013 at 20:37
• Why doesn't this work for any subgroup, not just one whose order is 1 less than the order of the group? Commented Feb 23, 2018 at 23:58
• The way I understand it intuitively is that we know $x*y = x$ for all $y \in H$, otherwise $H$ is not closed (this is where you need the order to be exactly 1 less.) Then multiply both sides of the equation by $x^{-1}$ on the left to obtain $1 = y$ for all $y \in H$, which is obviously a problem because we've got more than 1 element in $H$. Commented Oct 4, 2018 at 17:45

Consider an element $x$ in $G$ and not in $H$. Since $|H| = n - 1$, $x^{-1}$ must be in $H$. However, the inverse is unique so it's easy to see that $x^{-1}$ has no inverse in $H$ because $x$ is not in $H$.

• Cute! I'll give up +1 for cuteness! Commented Dec 3, 2013 at 20:13
• We might have $x^{-1} = x$ and this would fail. Commented Dec 3, 2013 at 20:15
• That is to say, this should not be voted up +1. Commented Dec 3, 2013 at 20:18
• @ Vladhagen: true enough! Which is I voted for cuteness! ;-) Ah well, as Treebeard the Ent put it, "Let us not ne hasty!" Commented Dec 3, 2013 at 23:13