Given a a subgroup $H$ of $G$ with index $3$, we have to show there is a subgroup $K$ of $G$ with index $2$, assuming that $H$ is not a normal subgroup of $G$.

My line of thinking was the following:

So $[G:H]=3$ and $H$ is not normal. This means there is a smaller prime which can divide $G$, thus, $2$ divides $G$. We then use Cauchy's theorem to say that there is an element $g$ in $G$ with order $2$.

Will it work to then use the permutation representation of left multiplication on $G$ and the existence of an odd permutation?

Any hint in the right direction will be much appreciated. Thanks.

  • 1
    $\begingroup$ Hello and welcome to math.stackexchange. Well-written question, and it's appreciated that you explain your line of thinking. The statement can indeed only be true if $H$ is not a normal subgroup, so this should be stated clearly in the question. I'll edit your post to refelct this. $\endgroup$ – Hans Engler Feb 16 '15 at 14:35

Let $G$ act on the cosets of $H$, we obtain a morphism $G \to S_3$ with Kernel $N$. It is well known that we have $N \leq H$. Since $H$ is not normal, $G \to S_3$ is surjective, because the image has more than $3$ elements. By the sign we obtain a surjection $G \to S_3 \to \{-1,1\}$, hence a subgroup of index $2$.


Hint: The action of $G$ on the left cosets of $H$ via left multiplication gives a homomorphism: $\phi: G \rightarrow S_{3}$. The $\ker \phi$ is the largest normal subgroup contained in $H$. Now consider $\frac{|G|}{|\ker \phi | } = \frac{3|H|}{|\ker \phi |} \in \{1, 2, 3, 6 \}$. Use the fact that $|\ker \phi|$ divides $|H|$, that $2$ does not divide $3$ and the fact that $H$ is not normal to deduce that $G$ would be $S_{3}$ which has a subgroup of index 2.


In general:

Proposition If $p$ is the smallest prime dividing the order of a group $G$, then a subgroup of index $p$ is normal in $G$.

The proof of this fact is along the same reasoning as the others here on this page: if $H$ is a subgroup of index $p$ let $G$ act on the right cosets of $H$ by right multiplication. The kernel of this action, called $core_G(H)$, is a normal subgroup of $G$ contained in $H$. Then $G/core_G(H)$ embeds homomorphically into $S_p$. And the order of $S_p$ is $p \cdot (p-1) \cdots 2 \cdot 1$.

Note (see remark of Marc van Leeuwen below, which is totally justified). I misread the OP post. Let's proceed along the lines of my reasoning above: we have $core_G(H) \subsetneq H \subsetneq G$. The first inclusion is strict, since $H$ is not normal. Since $G/core_G(H)$ embeds homomorphically into $S_3$, it follows that in fact $G/core_G(H) \cong S_3$. Since $A_3$ has index 2 in $S_3$, there has to be a subgroup $K$ of $G$, such that $core_G(H) \subseteq K$ and $K/core_G(H) \cong A_3$. But then index$[G:K]=2$.

  • $\begingroup$ Maybe you could also explain how the proposition answers the question. $\endgroup$ – Marc van Leeuwen Feb 16 '15 at 16:50

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.