# Minimal normal subgroups are elementary Abelian

• May 27th 2010, 06:34 AM
Giraffro
Minimal normal subgroups are elementary Abelian
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.
• May 27th 2010, 06:54 AM
Swlabr
Quote:

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 $H \lhd G$.
• May 27th 2010, 07:48 AM
Giraffro
Quote:

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.
• May 27th 2010, 08:05 AM
Swlabr
Quote:

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.
• May 27th 2010, 09:40 AM
Giraffro
Quote:

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!
• May 28th 2010, 12:34 AM
Swlabr
Quote:

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 $K \lhd H \lhd G \not\Rightarrow K \lhd G$.
• May 28th 2010, 12:40 AM
Giraffro
Quote:

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.
• May 28th 2010, 12:44 AM
Swlabr
Quote:

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...!
• May 28th 2010, 01:10 AM
Giraffro
Quote:

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!