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I recently read about the theorem that for a finite group $G$, if $p$ is the least prime dividing $|G|$, then any subgroup $H$ with $[G\colon H]=p$ is normal in $G$.

Going over some exercises, this theorem made it easy to see that a subgroup $H$, with $|H|=35$, is normal in $G$ when $|G|=70$.

A similar situation wasn't as obvious. If $|H|=35$, but $|G|=140$, then $H$ has index $4$. How then would one go about proving $H\unlhd G$? Thanks.

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Really, the theorem you quote is a very special case. In "most" circumstances, you aren't nearly so lucky as to have a subgroup of prime index, let alone a subgroup whose prime index is the smallest prime dividing $|G|$. So proving that a subgroup is normal from purely numerical considerations is usually difficult, if not impossible. (E.g., it would be impossible to prove that a subgroup of order $12$ in a group of order $60$ is normal). – Arturo Magidin Sep 1 '11 at 23:52
@Arturo, ah ok, that's too bad. I thought the approach may be similar. Is there another way to prove $H\unlhd G$ in this case using a different method? – yunone Sep 1 '11 at 23:53
When one can do it, it is usually through the application of the Sylow theorems or Hall's generalization. – Arturo Magidin Sep 1 '11 at 23:59
up vote 8 down vote accepted

A $7$-Sylow subgroup of $G$ has order $7$; the number of $7$-Sylow subgroups must be congruent to $1$ modulo $7$, and must divide $140 = 7\times 5\times 4$. Thus, it must divide $20$ and be congruent to $1$ modulo $7$; the only possibility is that the number of $7$-Sylow subgroups of $G$ is one (hence it is normal; in fact, characteristic) Thus, the subgroup $H$ of order $35$ must contain the unique $7$-Sylow subgroup of $G$.

Likewise, a $5$-Sylow subgroup of $G$ has order $5$, and the number of $5$-Sylow subgroups of $G$ must be congruent to $1$ modulo $5$ and divide $140$, hence must divide $7\times 4$. The only possibility is that there is a unique $5$-Sylow subgroup of $G$ (which is therefore characteristic, and thus normal), which must also be contained in $H$.

Thus, $H$ must be product of the unique $7$-Sylow subgroup and the unique $5$-Sylow subgroup of $G$. Since $H$ is the product of characteristic subgroups, it is in fact characteristic and therefore normal in $G$.

(Note that any group of order 35 is necessarily cyclic, by the by).

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Thanks Arturo. I just had two questions. Why does $H$ necessarily contain both the Sylow subgroups? And why does a $p$-Sylow subgroup being the unique one of its order imply it is characteristic? – yunone Sep 2 '11 at 1:18
@yunone: Since $H$ is of order $35$, it contains a subgroup of (of itself) of order $5$. This subgroup is also a subgroup of $G$, and is thus a $5$-Sylow subgroup of $G$ (it has the correct number of elements). But since $G$ has only one $5$-Sylow subgroup, then this subgroup of $H$ is the $5$-Sylow subgroup of $G$. Similar argument works for the $7$-Sylow subgroup. For your second question: if $K$ is a subgroup of $G$, and $K$ is the only subgroup of $G$ of order $|K|$, then $K$ is characteristic: for any $\phi\in\mathrm{Aut}(G)$, $\phi(K)$ is a subgroup of $G$ of order $|K|$. – Arturo Magidin Sep 2 '11 at 2:24
Oh whoops, I should have known that. Thanks for your help. – yunone Sep 2 '11 at 2:45

Since $[G:H]=4$, there is a homomorphism $\phi \colon G\rightarrow S_4$ with $ker(\phi)\subseteq H$. (See "Generalized Cayley's theorem- Introduction to theory of groups : Rotman )

If $ker(\phi)\neq H$, then it will be a subgroup (of $H$) of order $1,5$ or $7$.

  • If $|Ker(\phi)|=7$ then $|Im(\phi)|=|G/ker(\phi)|=140/7=20$; but $S_4$ has no subgroup of order 20.

  • If $|Ker(\phi)|\leq 5$ then $|Im(\phi)|=|G/ker(\phi)|\geq 140/5>24$, contradiction.

Therefore $ker(\phi)=H$ which is normal in $G$.

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Thanks group, I had been meaning to browse through Rotman's book soon. – yunone Sep 3 '11 at 3:07

I think I may have thought up an answer which doesn't use a whole lot of group theory. If anyone reads this, I'd appreciate feedback on whether it's right or wrong.

Let $H$ act on $G/H$ by $h\cdot gH=hgH$. Since $[G:H]=4$, the orbit of any $gH\in G/H$ is at most $4$. But the order of any orbit must divide $|H|$, so must be $1$, $5$, $7$, or $35$. So each orbit is of order 1, and thus trivial since $egH=gH$ in particular. So $hgH=gH$ for any $g\in G$, $h\in H$, hence $g^{-1}hgH=g^{-1}gH=H$. Then $g^{-1}hg\in H$, so $H\unlhd G$?

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You cannot go from $g^{-1}hgH$ to $g^{-1}gH$ in general, though you can go, here, to $H$ (since $hgH = gH$ is already known). $g^{-1}hgH = H$ suffices to show $g^{-1}hg\in H$. – Arturo Magidin Sep 8 '11 at 13:20
Thanks for the feedback. – yunone Sep 8 '11 at 14:07
@yunone: in other words, your last string of equalities might confuse the grader. Maybe better, "so dividing on the left by $g$, we get $g^{-1}hgH = g^{-1}gH$, but the right hand side is simply $H$. Hence $g^{-1}hg \in H$ and $H \unlhd G$." – Jack Schmidt Sep 8 '11 at 17:04
@Jack, ah ok, I was mostly just concerned on whether it was right or wrong, since I'm just learning. Thanks for pointing out what may be confusing to someone else, as the writer, I might not notice. – yunone Sep 8 '11 at 22:02

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