# Without using Sylow: Group of order 28 has a normal subgroup of order 7

Prove that a group of order 28 has a normal subgroup of order 7.

How can I prove this without using Sylow's theorem?

I know by Cauchy’s theorem, there exists an $x\in G$ with order 7, now I just need to prove it has a normal subgroup.

• Hint: $G$ acts on the cosets of this subgroup by translation, giving a homomorphism to $S_4$ with transitive image. What is the kernel of this? Nov 11, 2013 at 10:09

Another different proof: since as you said you know that $$G$$ has an element of order $$7$$, by Cauchy, you also know that there's at least one subgroup of order $$7$$, let's call it $$H$$.

Suppose that $$K \leq G$$ is another subgroup of order $$7$$, then we can consider the subset $$HK$$ that has order $$|HK|=|H||K|/|H \cap K|$$.

If $$H$$ and $$K$$ were distinct then $$H \cap K$$ should be a proper subgroup of both of them, but since they have order the prime $$7$$ this is possible iff $$H \cap K=(\text{id})$$ and so $$|HK| =7\cdot 7 = 49$$ which is clearly bigger then $$28$$.

We arrived to an absurdity, so we have to conclude that $$H$$ is the only subgroup of order $$7$$ and so it's characteristic, hence normal.

Edit (more details): let's consider a generic automorphism $$\varphi \colon G \to G$$, then by properties of homomorphisms $$\varphi(H)$$ is a subgroup of $$G$$ and since $$\varphi$$ is bijective $$\varphi(H)$$ should have order $$7$$.

Because as we have proved $$H$$ is the only subgroup of order $$7$$ it follows that $$\varphi(H)=H$$: so $$H$$ is fixed by all the automorphisms, i.e. is characteristic.

From this follows normality since a subgroup is normal iff is fixed by all inner automorphisms, i.e. is fixed by all the automorphisms of the form $$x \mapsto gxg^{-1}$$ for some $$g \in G$$.

Since $$H$$ is fixed by every automorphism it's fixed in particular by the inner automorphism and so it's normal.

• I didn't understand the last part. How did you arrive that it is normal since H is the only subgroup of order 7? Nov 13, 2013 at 3:32
• @user104235 I've added details, tell me if you have any other problem :) Nov 13, 2013 at 8:50
• The formulation of your "more details" part is slightly complicated, in my opinion. See the Sylow part of my answer for a shorter alternative. BTW, I really like the first part of your (+1) answer. Nice argument! Nov 13, 2013 at 21:30
• @azimut what's more complicated? I've just proved that a subgroup which is the only one of its order is characteristic and that a characteristic subgroup is normal. It's really that bad? Nov 13, 2013 at 21:32
• No, your argument is perfectly fine. But no need to talk about general automorphisms, characteristic subgroups etc. What about "Let $g\in G$. Since conjugation is an automorphism, $gHg^{-1}$ is again a subgroup of order $7$. And since $H$ is the only such subgroup, $gHg^{-1} = H$. So $H$ is normal in $G$." Nov 13, 2013 at 21:41

We have $\lvert G\rvert = 28 = 2^2\cdot 7$. Let $a_7$ be the number of $7$-Sylow groups in $G$. By Sylow $$a_7\equiv 1\mod 7\qquad\text{and}\qquad a_7 \mid 4\text{.}$$ This implies $a_7 = 1$, so there is a unique subgroup $H$ of $G$ of order $7$.

For all $g\in G$, $gHg^{-1}$ is again a subgroup of order $7$, which forces $gHg^{-1} = H$ for all $g\in G$. So $H$ is a normal subgroup of order $7$.

Edit Only now I realized that you don't want to use the Sylow theorems. Here is an alternative, following the hint of @Tobias Kildetoft:

Since $7$ is prime, by Cauchy $G$ contains an element of order $7$. It generates a subgroup $H$ of $G$ of order $7$. Look at the group action of $G$ on the set $G/H$ of left-cosets of $H$ by left multiplication. Obviously, this group action is transitive. Because of $\lvert G/H\rvert = 28/7 = 4$, it gives rise to a group homomorphism $$\varphi : G \to S_4$$ Now look at $N = \ker(\varphi)$, which is a normal subgroup of $G$. From the transitivity of $\varphi$, we get $\operatorname{im}(\varphi) \geq 4$, which translates to $\lvert N\rvert \leq 28/4 = 7$.

By the homomorphism theorem, $G/N \cong \operatorname{im}(\varphi)$, so $\lvert G\rvert/\lvert N\rvert$ divides $\lvert S_4\rvert = 24$. This implies $28 \mid 24 \lvert N\rvert$, so $7\mid\lvert N\rvert$.

So the only remaining possitiblity is $\lvert N\rvert = 7$.

• The OP wants to avoid Sylow.
– lhf
Nov 11, 2013 at 11:15
• @lhf: Yeah, I just realized that, too. So I added an alternative without Sylow. Nov 11, 2013 at 11:16
• You forgot the possibility of $|N| = 14$, which is where transitivity is really needed (otherwise, non-trivial would have been sufficient). Nov 11, 2013 at 11:47
• @TobiasKildetoft: Right, thank you. I've modified my answer. Nov 11, 2013 at 11:53
• The order of $S_4$ is $4!=24$.
– user45861
Nov 13, 2013 at 9:14

Another kind of proof. First, we remark that if $G$ is abelian, we can apply Cauchy to get a group of order $7$, which must be normal.

Consider the conjugacy classes of a non-abelian $G$. They can only have sizes $1,2,4,7,14,28$. We know the identity is in a class by itself, so we have $27$ more elements to divvy up. We would like to show one of them has conjugacy class $1$, $2$, or $4$.

Clearly, we cannot form $27$ out of just $7$s and $14$s, so at least one of those three is required, call it $k$. $G$ acts on the elements of that conjugacy class by conjugation, so it induces a homomorphism $\varphi$ from $G$ into $S_k$. But $|S_k|$ is at most $24$, which is less than $28$, so $\ker \varphi$ is non-trivial. Since $G$ is non-abelian, $\ker \varphi \ne G$, so we have our normal subgroup, $\ker \varphi$.

This works for groups of order $mn$, where $m$ and $n$ are relatively prime and $n \nmid (m -1)!$.

• I don't see why $\ker\varphi$ is of order $7$. In the case $k = 2$, wouldn't we get $\lvert\ker(\varphi)\rvert = 14$? Nov 13, 2013 at 11:25
• Oh, right, we would. Hmm. There is a subgroup of order $7$ normal in the kernel, but normality isn't transitive. I'm not actually sure what to do from here. :\ Nov 13, 2013 at 18:53