Let $G$ be a group of odd order, and let $H$ be a subgroup of $G$ of index $5$. Prove that $H$ is a normal subgroup of $G$.

Question

Let $G$ be a group of odd order, and let $H$ be a subgroup of $G$ of index $5$. Prove that $H$ is a normal subgroup of $G$.

Hint: Show that there is no subgroup of order $15$ in $S_5$.

I did prove the hint, but don't see how it helps.

Answer 1

Here is a series of hint:

-Assume that $H$ is not normal

-Argue that $H$ is self-normalising in $G$

-Conclude that $H$ has exactly $5$ conjugates in $G$

-Use this to construct a homomorphism from $G$ to a transitive subgroup of $S_5$

-Using the hint, determine what is the image of this homomorphism

-Derive a contradiction

Answer 2

In addition to the things answered here, I just want to add this one.

Proposition (Van Der Waall, Bioch) Let $G$ be a finite group and $H$ a subgroup of prime index $p$, with gcd$(|G|,p-1)=1$. Then $G' \subseteq H$.

Note that this implies that $H \unlhd G$, and that it is in fact sufficient to prove that $H$ is normal, since then $G/H \cong C_p$ is abelian.

Proof Firstly, we may assume by induction on $|G|$, that $H$ is core-free, that is core$_G(H)=\bigcap_{g \in G}H^g=1$. This means that $G$ can be homomorphically embedded in $S_p$. Let $P \in Syl_p(G)$ and note that because $|S_p|=p \cdot (p-1) \cdots \cdot 1$, $|P|=p$. By the $N/C$-Theorem, $N_G(P)/C_G(P)$ embeds in Aut$(P) \cong C_{p-1}$. By the assumption gcd$(|G|,p-1)=1$, we get that $N_G(P)=C_G(P)$. Since $P$ is abelian we have $P \subseteq C_G(P)$, whence $P \subseteq Z(N_G(P))$. We now can apply Burnside's Normal $p$-complement Theorem, which implies that $P$ has a normal complement $N$, that is $G=PN$ and $P \cap N=1$. Note that $|G/N|=p$.

Look at the image of $H$ in $G/N$. Then $G=HN$, or $HN=N$. In the latter case $H \subseteq N$, and $|G:H|=|G:N|=p$, whence $H=N$ and we are done if we can refute the first case. If $G=HN$, then $|G:H \cap N|=|G:N|\cdot|N:H \cap N|=|G:N|\cdot |G:H|=p \cdot p=p^2$, contradicting the fact that $|G| \mid |S_p|$. The proof is now complete.

Corollary 1 Let $G$ be a finite group and let $H$ be a subgroup with $|G:H|=p$, the smallest prime dividing the order of $G$. Then $G' \subseteq H$. In particular, $H$ is normal.

Corollary 2 Let $G$ be a finite group and let $H$ be a subgroup with $|G:H|=p$ and gcd$(|H|,p-1)=1$. Then $H$ is normal.

Observe that this last result renders a well-known result for $p=2$! Finally for fun:

Corollary 3 Let $G$ be a finite group of odd order and $H$ a subgroup with $|G:H|=65537$ (or any Fermat prime). Then $H$ is normal.

Answer 3

Let $H$ act on the left cosets of $H$ by left multiplication (I.e., $h*(xH)=(hx)H$.) Consider the orbits of this action. No orbit can have even size as the orbit must divide the order of G, which is odd. Therefore the orbit sizes are 1, 3, or 5. Clearly H has orbit 1, so this means either the orbits are all size 1, or there are two orbits of size 1 and one orbit of size 3. In either case, there is another orbit gH of size 1, which implies g normalises H, which means that the normaliser of H is strictly a super group of H. But then Lagrange implies that the normaliser of H is exactly G. This means H is normal.

Answer 4

Here is another solution, which uses somewhat more "classical" results than the one used by Nicky Hekster.

Let $G$ be a group of odd order, and $H$ a subgroup of $G$ of index $5$. By Cayley's theorem, there is a homomorphism $\varphi:G\to S_5$, whose kernel, say, $K$, is contained in $H$. Let $\overline{G}$ denote the image of $G$ in $S_5$ under this homomorphism. We have $$|\overline{G}|=[G:K]=[G:H][H:K]=5[H:K]$$ Hence $5[H:K]$ divides $\overline{G}$ which divides $5!=120$, so $[H:K]$ divides $4!=24$, so it must be one of the numbers $\{1,2,3,4,6,8,12\}$. On the other hand, we also have $|G|=|\overline{G}|\cdot |K|$, so $|\overline{G}|$ divides $|G|$, and hence $5[H:K]$ also divides $|G|$. By assumption, $[H:K]$ must be odd, or else $|G|$ would be even, contradicting the assumption. Looking at the list of possible candidates for $[H:K]$, the only odd options are $[H:K]=1$ or $[H:K]=3$. Assume the latter. Then $|\overline{G}|=15$. This implies that $S_5$ has a subgroup of order $15$. But every group of order $15$ is cyclic, so $S_5$ must have an element of order $15$. But that is a contradiction, because the order of a permutation is equal to the least common multiple of the orders of the disjoint cycles that represent it, and it is impossible to form disjoint cycles whose lengths are $5$ and $3$ as permutations on $5$ elements only. This contradiction shows that the only option for $[H:K]$ is $[H:K]=1$, but $K$ is a subgroup of $H$ so $H=K$, and $K$ is a kernel of a homomorphism, so it is normal in $G$.