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.

  • 8
    $\begingroup$ 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? $\endgroup$ Nov 11, 2013 at 10:09

3 Answers 3


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.

  • $\begingroup$ I didn't understand the last part. How did you arrive that it is normal since H is the only subgroup of order 7? $\endgroup$
    – user104235
    Nov 13, 2013 at 3:32
  • $\begingroup$ @user104235 I've added details, tell me if you have any other problem :) $\endgroup$ Nov 13, 2013 at 8:50
  • $\begingroup$ 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! $\endgroup$
    – azimut
    Nov 13, 2013 at 21:30
  • $\begingroup$ @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? $\endgroup$ Nov 13, 2013 at 21:32
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    $\begingroup$ 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 $G$." $\endgroup$
    – azimut
    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 $\lvert N\rvert = 7$.

  • $\begingroup$ The OP wants to avoid Sylow. $\endgroup$
    – lhf
    Nov 11, 2013 at 11:15
  • $\begingroup$ @lhf: Yeah, I just realized that, too. So I added an alternative without Sylow. $\endgroup$
    – azimut
    Nov 11, 2013 at 11:16
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    $\begingroup$ You forgot the possibility of $|N| = 14$, which is where transitivity is really needed (otherwise, non-trivial would have been sufficient). $\endgroup$ Nov 11, 2013 at 11:47
  • $\begingroup$ @TobiasKildetoft: Right, thank you. I've modified my answer. $\endgroup$
    – azimut
    Nov 11, 2013 at 11:53
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    $\begingroup$ The order of $S_4$ is $4!=24$. $\endgroup$
    – 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 $n \nmid (m -1)!$.

  • $\begingroup$ I don't see why $\ker\varphi$ is of order $7$. In the case $k = 2$, wouldn't we get $\lvert\ker(\varphi)\rvert = 14$? $\endgroup$
    – azimut
    Nov 13, 2013 at 11:25
  • $\begingroup$ 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. :\ $\endgroup$ Nov 13, 2013 at 18:53

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