Minimal normal subgroups are elementary Abelian

**Giraffro** · May 27th 2010, 06:34 AM

Hey, can anybody help me with this?

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

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

**Swlabr** · May 27th 2010, 06:54 AM

Originally Posted by Giraffro

Hey, can anybody help me with this?

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

I have no idea where to start with this. I can see that all proper non-trivial subgroups of $M$ aren't normal in $G$ , 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 $H$ is a characteristic subgroup of $K$ and $K \lhd H$ then $H \lhd G$ .

**Giraffro** · May 27th 2010, 07:48 AM

Originally Posted by Swlabr

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 $H$ is a characteristic subgroup of $K$ and $K \lhd H$ then $H \lhd G$ .

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** · May 27th 2010, 08:05 AM

Originally Posted by Giraffro

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,

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

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** · May 27th 2010, 09:40 AM

Originally Posted by Swlabr

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,

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

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

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

Thanks for the help!

**Swlabr** · May 28th 2010, 12:34 AM

Originally Posted by Giraffro

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

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

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

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

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

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

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

**Giraffro** · May 28th 2010, 12:40 AM

Originally Posted by Swlabr

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

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

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

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

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

I think there's a typo in my proof: I meant to say that $\phi(P) \in \text{Syl}_p(M)$ , since $M \lhd G$ and as you say, automorphisms preserve group order.

**Swlabr** · May 28th 2010, 12:44 AM

Originally Posted by Giraffro

I think there's a typo in my proof: I meant to say that $\phi(P) \in \text{Syl}_p(M)$ , since $M \lhd G$ 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** · May 28th 2010, 01:10 AM

Originally Posted by Swlabr

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!

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