# Minimal normal subgroups are elementary Abelian

#### Giraffro

Hey, can anybody help me with this?

If $$\displaystyle G$$ is a finite group with a minimal non-trivial normal subgroup $$\displaystyle M$$ such that $$\displaystyle M$$ is Abelian, then $$\displaystyle \exists p \in \mathbb{N}^+$$ prime such that $$\displaystyle \forall x \in M \backslash \{e\}, o(x) = p$$.

I have no idea where to start with this. I can see that all proper non-trivial subgroups of $$\displaystyle M$$ aren't normal in $$\displaystyle G$$, but not sure how this helps.

#### Swlabr

Hey, can anybody help me with this?

If $$\displaystyle G$$ is a finite group with a minimal non-trivial normal subgroup $$\displaystyle M$$ such that $$\displaystyle M$$ is Abelian, then $$\displaystyle \exists p \in \mathbb{N}^+$$ prime such that $$\displaystyle \forall x \in M \backslash \{e\}, o(x) = p$$.

I have no idea where to start with this. I can see that all proper non-trivial subgroups of $$\displaystyle M$$ aren't normal in $$\displaystyle G$$, but not sure how this helps.
You have two things to prove, that only one prime divides the order of your subgroup and that your subgroup is a direct product of groups of order p.

Hint: If $$\displaystyle H$$ is a characteristic subgroup of $$\displaystyle K$$ and $$\displaystyle K \lhd H$$ then $$\displaystyle H \lhd G$$.

• Giraffro

#### Giraffro

You have two things to prove, that only one prime divides the order of your subgroup and that your subgroup is a direct product of groups of order p.

Hint: If $$\displaystyle H$$ is a characteristic subgroup of $$\displaystyle K$$ and $$\displaystyle K \lhd H$$ then $$\displaystyle H \lhd G$$.
I haven't encountered characteristic subgroups before and this question is from a past paper a few years ago, so I may just skip it. Thanks for replying anyway.

#### Swlabr

I haven't encountered characteristic subgroups before and this question is from a past paper a few years ago, so I may just skip it. Thanks for replying anyway.
I'm sure there are other ways, but that way seemed the neatest. Basically, a characteristic subgroup is a subgroup which is fixed under automorphisms,

$$\displaystyle \phi : G \rigtharrow G$$ then $$\displaystyle H\phi = H$$.

Examples of characteristic subgroups are the two boring subgroups, the center, the commutator subgroup, and the frattini subgroup.

If you take a cyclic group, then every subgroup is characteristic.

Now, you can think of conjugation by an element as an automorphism, and so such groups are normal. However, the characteristic subgroup of a normal subgroup will be normal in the big group, and this is what I would have utilised.

For the first part, if another prime divided the group M then take the largest q-group in M. As your group is abelian this will encompass all the q-ness of your group (q will not divide the order of M/Q). Therefore, you would have a characteristic subgroup, which will thus be normal.

• Giraffro

#### Giraffro

I'm sure there are other ways, but that way seemed the neatest. Basically, a characteristic subgroup is a subgroup which is fixed under automorphisms,

$$\displaystyle \phi : G \rigtharrow G$$ then $$\displaystyle H\phi = H$$.

Examples of characteristic subgroups are the two boring subgroups, the center, the commutator subgroup, and the frattini subgroup.

If you take a cyclic group, then every subgroup is characteristic.

Now, you can think of conjugation by an element as an automorphism, and so such groups are normal. However, the characteristic subgroup of a normal subgroup will be normal in the big group, and this is what I would have utilised.

For the first part, if another prime divided the group M then take the largest q-group in M. As your group is abelian this will encompass all the q-ness of your group (q will not divide the order of M/Q). Therefore, you would have a characteristic subgroup, which will thus be normal.
So if I understand you correctly, this is how you prove it:

Let $$\displaystyle p \in \mathbb{N}^+$$ be prime such that $$\displaystyle p \mid |M|$$ and let $$\displaystyle P \in \text{Syl}_p(G)$$ be given. Then since $$\displaystyle M$$ is Abelian, $$\displaystyle P \lhd M$$ and so $$\displaystyle |\text{Syl}_p(G)| = 1$$. Now let $$\displaystyle g \in G$$ be given and define $$\displaystyle \phi : G \to G, h \mapsto h^g$$. Then $$\displaystyle \phi \in \text{Aut}(G)$$ and since $$\displaystyle M \lhd G, \phi|_M \in \text{Aut}(M)$$. Thus $$\displaystyle \phi(P) \in \text{Syl}_p(G)$$ and so by uniqueness $$\displaystyle P = \phi(P) = P^g$$. Therefore $$\displaystyle P \lhd G$$ and since $$\displaystyle M$$ is a minimal non-trivial normal subgroup of $$\displaystyle G, P = M$$ or $$\displaystyle P = \{e\}$$. However, $$\displaystyle p \mid |M|$$, so by Sylow's theorems, $$\displaystyle p \mid |P|$$ and therefore $$\displaystyle P = M$$. Hence $$\displaystyle M$$ is an Abelian $$\displaystyle p$$-group.

Finally, define $$\displaystyle H := \{g \in M : o(g) \mid p\}$$. Then $$\displaystyle e \in H$$ and as $$\displaystyle M$$ is Abelian, $$\displaystyle \forall g, h \in M, o(gh) = lcm(o(g), o(h)) \mid p$$ and $$\displaystyle o(g^{-1}) = o(g) \mid p$$, so $$\displaystyle gh, g^{-1} \in H$$. Therefore $$\displaystyle H \lhd M$$. Now let $$\displaystyle g \in G, h \in H$$ be given. Then $$\displaystyle o(h^g)=o(h) \mid p$$, so since $$\displaystyle M \lhd G, h^g \in M$$ and therefore $$\displaystyle h^g \in H$$. Thus $$\displaystyle H \lhd G$$ and so $$\displaystyle H = M$$ or $$\displaystyle H = \{e\}$$. However, by Cauchy's theorem $$\displaystyle \exists g \in M$$ such that $$\displaystyle o(g) = p$$ and thus $$\displaystyle g \in H \backslash \{e\}$$. Therefore $$\displaystyle M = H$$.

Thanks for the help!

#### Swlabr

So if I understand you correctly, this is how you prove it:

Let $$\displaystyle p \in \mathbb{N}^+$$ be prime such that $$\displaystyle p \mid |M|$$ and let $$\displaystyle P \in \text{Syl}_p(G)$$ be given. Then since $$\displaystyle M$$ is Abelian, $$\displaystyle P \lhd M$$ and so $$\displaystyle |\text{Syl}_p(G)| = 1$$. Now let $$\displaystyle g \in G$$ be given and define $$\displaystyle \phi : G \to G, h \mapsto h^g$$. Then $$\displaystyle \phi \in \text{Aut}(G)$$ and since $$\displaystyle M \lhd G, \phi|_M \in \text{Aut}(M)$$. Thus $$\displaystyle \phi(P) \in \text{Syl}_p(G)$$ and so by uniqueness $$\displaystyle P = \phi(P) = P^g$$. Therefore $$\displaystyle P \lhd G$$ and since $$\displaystyle M$$ is a minimal non-trivial normal subgroup of $$\displaystyle G, P = M$$ or $$\displaystyle P = \{e\}$$. However, $$\displaystyle p \mid |M|$$, so by Sylow's theorems, $$\displaystyle p \mid |P|$$ and therefore $$\displaystyle P = M$$. Hence $$\displaystyle M$$ is an Abelian $$\displaystyle p$$-group.
Not quite...you are taking a sylow subgroup of M, not of G. There is no reason to say these coincide.

Let $$\displaystyle p | |M|$$. As $$\displaystyle M$$ is abelian, the sylow $$\displaystyle p$$-subgroups of $$\displaystyle M$$ are normal and unique, so let $$\displaystyle P$$ be the maximal subgroup of $$\displaystyle M$$ of order a power of $$\displaystyle p$$. Clearly, if $$\displaystyle |M| = p^nq$$, $$\displaystyle p \not\div q$$ then $$\displaystyle |P| = p^n$$.

Now, we know that $$\displaystyle P \lhd M$$, but we wish to prove that $$\displaystyle P \lhd G$$. This is true because if $$\displaystyle \phi: M \rightarrow M$$ is an automorphism, then $$\displaystyle P\phi = P$$ because every element has order a power of $$\displaystyle p$$ and no other element in the group has this order, and automorphisms preserve order (does that make sense)? Therefore, $$\displaystyle P \text{ char } M$$.

Next, notice that for $$\displaystyle g \in G$$, $$\displaystyle ^g:M \rightarrow M$$, $$\displaystyle h \mapsto h^g$$ is an automorphism of $$\displaystyle M$$. This is bacause $$\displaystyle M$$ is normal in $$\displaystyle G$$. As this is an automorphism of $$\displaystyle M$$ it must fix $$\displaystyle P$$. Therefore, $$\displaystyle P^g = P$$ for all $$\displaystyle g \in G$$, and we thus conclude that $$\displaystyle P \lhd G$$, as required to get our contradiction.

Note that $$\displaystyle K \lhd H \lhd G \not\Rightarrow K \lhd G$$.

#### Giraffro

Not quite...you are taking a sylow subgroup of M, not of G. There is no reason to say these coincide.

Let $$\displaystyle p | |M|$$. As $$\displaystyle M$$ is abelian, the sylow $$\displaystyle p$$-subgroups of $$\displaystyle M$$ are normal and unique, so let $$\displaystyle P$$ be the maximal subgroup of $$\displaystyle M$$ of order a power of $$\displaystyle p$$. Clearly, if $$\displaystyle |M| = p^nq$$, $$\displaystyle p \not\div q$$ then $$\displaystyle |P| = p^n$$.

Now, we know that $$\displaystyle P \lhd M$$, but we wish to prove that $$\displaystyle P \lhd G$$. This is true because if $$\displaystyle \phi: M \rightarrow M$$ is an automorphism, then $$\displaystyle P\phi = P$$ because every element has order a power of $$\displaystyle p$$ and no other element in the group has this order, and automorphisms preserve order (does that make sense)? Therefore, $$\displaystyle P \text{ char } M$$.

Next, notice that for $$\displaystyle g \in G$$, $$\displaystyle ^g:M \rightarrow M$$, $$\displaystyle h \mapsto h^g$$ is an automorphism of $$\displaystyle M$$. This is bacause $$\displaystyle M$$ is normal in $$\displaystyle G$$. As this is an automorphism of $$\displaystyle M$$ it must fix $$\displaystyle P$$. Therefore, $$\displaystyle P^g = P$$ for all $$\displaystyle g \in G$$, and we thus conclude that $$\displaystyle P \lhd G$$, as required to get our contradiction.

Note that $$\displaystyle K \lhd H \lhd G \not\Rightarrow K \lhd G$$.
I think there's a typo in my proof: I meant to say that $$\displaystyle \phi(P) \in \text{Syl}_p(M)$$, since $$\displaystyle M \lhd G$$ and as you say, automorphisms preserve group order.

#### Swlabr

I think there's a typo in my proof: I meant to say that $$\displaystyle \phi(P) \in \text{Syl}_p(M)$$, since $$\displaystyle M \lhd G$$ and as you say, automorphisms preserve group order.
Oh - it's clearly a typo too. I should have read your proof through properly. It would have saved about 15 minutes of typing...!

#### Giraffro

Oh - it's clearly a typo too. I should have read your proof through properly. It would have saved about 15 minutes of typing...!
No worries. Thanks for the help!