Minimal normal subgroups are elementary Abelian

Mar 2010
36
9
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.
 
May 2009
1,176
412
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 \(\displaystyle H \lhd G\).
 
  • Like
Reactions: Giraffro
Mar 2010
36
9
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.
 
May 2009
1,176
412
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.
 
  • Like
Reactions: Giraffro
Mar 2010
36
9
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!
 
May 2009
1,176
412
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 \(\displaystyle K \lhd H \lhd G \not\Rightarrow K \lhd G\).
 
Mar 2010
36
9
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.
 
May 2009
1,176
412
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...!