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This is my question : Let G be a group without subgroups with index 2 (i.ee. there is no subgroup M s.t $[G:M]=2$).prove that every group $H<G$ s.t $[G:H]=3$ is normal in G.

What I thought to do: Let N be a subgroup s.t $[G:H]=3$ so $G/H={N,xN,yN}$ so every $g \in G$ is of the form : $g=n,g=xn,g=yn$ when $n \in \mathbb N$. I tried to prove that for every $g \in G$ : $gNg^{-1} =N$ but without a success.

Martin Sleziak
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Tom
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  • Ohhh I see now that G/H is cyclic, maybe it can help – Tom Feb 02 '15 at 13:50
  • http://math.stackexchange.com/questions/164244/normal-subgroup-of-prime-index – Jack Yoon Feb 02 '15 at 13:53
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    @JackYoon But that still leaves the case where the order is divisible by $2$. – Tobias Kildetoft Feb 02 '15 at 13:55
  • Maybe one should also point out, that the assertion does not carry over to the smallest prime $p$ of which a subgroup of index $p$ exists (in contrast to the assertion in the link above, where we consider the smallest prime, which divides the group order).

    $A_5$ contains no subgroup index $2$ and $3$, but a subgroup of index $5$ (namely $A_4$), that is not normal.

     – MooS Feb 02 '15 at 14:25
  • @user1729: please read the comments of Tobias ,it is not a duplicate. – mesel Feb 02 '15 at 21:49

2 Answers2

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Let $G$ act on the left coset of $H$ by left multiplication. Then we have an homomorphism $\phi$ from $G$ to $S_3$. As $G$ has no subgroup of index $2$, $\phi(G)\leq A_3$.

Notice that $\phi$ can not be trivial thus, $\phi(G)=A_3$. Then clearly $\ker \phi=H$.

Martin Sleziak
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mesel
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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 $N \leq H$, we need to prove equality. If not, then we have $[G:N]= 6$ and $G \to S_3$ is surjective. By the sign we obtain a surjection $G \to S_3 \to \{-1,1\}$, hence a subgroup of index $2$.

MooS
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