Any group of order 952 contains a subgroup of order 68?

[Syrus]

Homework Statement

I am struggling with a proof for this. Obviously Sylow's theorems come into play. We have that |G| = 952. As sylow's first theorem only covers subgroups of order pⁿ, we cannot directly use it to assert the existence of a subgroup of order 68. On the other hand, if we can come up with a (normal) subgroup H of G such that [G : H] = 14, then by lagranges theorem we would have that |H| = 68. I am just unsure of how to begin with this construction.

Homework Equations

The Attempt at a Solution

 

[jgens]

Start by using the Sylow Theorems to prove that there is a unique subgroup N of order 17.

Edit: My first response gave too much of the solution away. The main idea was to think about internal semi-direct products.

 

[Syrus]

You can use Sylow's first theorem to show that a sylow p-subgroup of order 17 exists. Then sylow's third theorem shows that it is unique. Finally, Sylow's second theorem asserts that it is normal in G. Now...

 

[jgens]

If we are thinking about constructing the subgroup of order 68 via an internal semi-direct product and we have a normal subgroup of order 17, then what kind of group should we be looking for now?

 

[Syrus]

I apologize jgens, I don't recall my abstract algebra class going into depth discussing semi-direct products. While I do think I could grasp them (after looking them up online) I don't want my instructor to think I simply copied a proof I don't understand. Is there an alternative proof strategy?

 

[jgens]

I apologize jgens, I don't recall my abstract algebra class going into depth discussing semi-direct products. While I do think I could grasp them (after looking them up online) I don't want my instructor to think I simply copied a proof I don't understand. Is there an alternative proof strategy?

Off the top of my head, I cannot think of an alternative way of proving this result without appealing to semi-direct products in some form. The problem is that most of the relevant theorems giving us information about the subgroups of G concern p-groups, while we want to find a subgroup which is not a p-group. The most obvious way of doing this is by taking two p-groups and combining them in some way to form another subgroup of G with the desired properties. And the easiest way of combining these groups is via semi-direct products.

In any case, you can probably prove the following theorem without too much difficulty: If $G$ is a group, $N,K \leq G$ are subgroups of $G$ and $N$ is normal in $G$, then $NK = \{nk \in G:n \in N, k \in K\}$ is a subgroup of $G$. The idea is to essentially use this result to build the group you want.

 

[Syrus]

I received this hint:

The key result is the fact that if N is normal in G, then there is
a 1-1 correspondence between subgroups of G containing N and subgroups
of the factor group G/N. If H is a subgroup of G containing N, then
the corresponding subgroup of G/N is H/N. So, you also have a simple
relationship between the orders of the subgroups in the above 1-1
correspondence (i.e. just multiply or divide by |N|). So, knowledge
of subgroups in G/N can tell you about subgroups of G.

Now..You can determine which subgroups of G are normal in G by Sylow's theorem. How do you determine the correct H containing N such that |H| = 68?

 

[jgens]

Hmm ... I didn't think about using the lattice isomorphism theorem, but this proof is essentially the same as the proof I was suggesting (although the lattice isomorphism theorem takes care of the semi-direct product). In any case, you want to find a subgroup of G/N with order 4. Can you figure out why?

 

[Syrus]

Use the sylow theorems to show that the sylow 17-subgroup (call it K) is unique and hence normal in G. Hence, G/K is a group of order [G : K] = 56. But 56 = (2²)(14), and so by Sylow's first theorem, G/K contains a subgroup of order 4.

How should I integrate the lattice isomorphism theorem here? I know all groups of order 4 are abelian and hence normal, but I was unsure as to whether this is the right track.

 

[jgens]

It's not quite that simple: Notice that 56 = 2³7 so Sylow's Theorem actually gives you a subgroup of order 8, not order 4. An abelian subgroup is also not necessarily normal (it is a good exercise to find an example of an abelian subgroup which is not normal).

In any case, once you correctly find a subgroup of G/N or order 4, then just apply the lattice isomorphism theorem directly to get your group of order 68.

 

[Syrus]

I'm trying to figure out how to construct subgroups of factor groups

 

[jgens]

Well start by using Sylow's Theorem to get a subgroup of G/N of order 8. Now for any group of order 8, can you find a subgroup of order 4?

 

[Syrus]

We've already shown above that a normal subgroup N of order 17 exists in G. Since G/N is then a group of order 56, we reapply sylow's theorems in a similar fashion to show that there exists a normal subgroup of G/N (which has the form L/N) having order 8. Applying Sylow's first theorem to L/N implies that it contains a subgroup K/N of order 2² = 4. By transitivity of subgroups, then, K/N ≤ G/N. Thus, by the lattice isomorphism theorem, there is a corresponding subgroup K of G of order [K : N]|N| = (4)(17) = 68.

This seems more valid than previous attempts.

 

[jgens]

we reapply sylow's theorems in a similar fashion to show that there exists a normal subgroup of G/N (which has the form L/N) having order 8.

If you want to talk about normality, then you should prove it. I don't know if there is always a normal subgroup of order 8, so it might not be possible. Luckily, it doesn't matter, so you don't need to mention it.

Applying Sylow's first theorem to L/N implies that it contains a subgroup K/N of order 2² = 4.

I recommend you read through the Sylow theorems again. None of the Sylow theorems allow you to conclude this.

 

[Syrus]

My text has:

Sylow's first theorem: Let pⁿ divide |G|. Then G has a subgroup of order pⁿ.

 

[Syrus]

I checked another book... it can't be the case that p divides m, where |G| = pⁿm. Strange that the first book states it that way then.

 

[jgens]

The first theorem is usually stated as: Let G be a group and suppose that |G| = pⁿm where p does not divide m. Then G has a subgroup of order pⁿ.

If there is a theorem which says "If pⁿ divides |G|, then G has a subgroup of order pⁿ" this is a stronger statement than the first Sylow Theorem (at least as its usually stated). I am actually not convinced this stronger statement is true and I don't really have time to think through it. But if you can use that result, then the proof is fine.

Edit: Does the first book state that G has a p-Sylow subgroup or Sylow p-subgroup? Because both of those terms mean that p cannot divide m.

 

[Syrus]

Every group of order 8 has a subgroup of order two (isomorphic to Z₂). Perhaps we can form the direct product of this group of order 2 to achieve our group of order 4?

 

[jgens]

Every group of order 8 has a subgroup of order two (isomorphic to Z₂). Perhaps we can form the direct product of this group of order 2 to achieve our group of order 4?

You are working too hard. There are five groups of order 8: $Z_8,Z_4 \times Z_2,Z_2 \times Z_2 \times Z_2,D_8,Q$. Each of these has a subgroup of order 4 (you should prove this).

Edit: You should do this without appealing to the Sylow Theorems.

 

[Syrus]

Jgens, once I prove that all groups of order 8 have a subgroup of order 4, I can place this in the proof and use the lattice isomorphism theorem to show that a subgroup H of order 68 exists in G.Before I do this however, I figure I pose another question as to the validity of one of my proofs before it gets too late:

Show that the extension field Q(√2,√3) of the rationals is equal to the simple extension Q(√2 + √3).

Proof: Clearly Q(√2,√3) is a subset of Q(√2 + √3) as Q(√2,√3) contains both √2 and √3 and hence also √2 + √3 by the field operations. Now, for nonzero p, we can divide p(√2 + √3) by p to obtain elements of the form q + √2 + √3, where of course the field operations allow √2 and √3 to be multiplied by any element of Q. Note too that(√2 + √3)(√2 + √3) = 2 + 2√2√3 + 3 which is also clearly expressible in Q(√2,√3) as √6 = √2√3.

 

[jgens]

Proof: Clearly Q(√2,√3) is a subset of Q(√2 + √3) as Q(√2,√3) contains both √2 and √3 and hence also √2 + √3 by the field operations.

You have the obvious inclusion backwards. It follows that $\mathbb{Q}(\sqrt{2}+\sqrt{3}) \leq \mathbb{Q}(\sqrt{2},\sqrt{3})$ for the reason you mention above. The trickier part is showing the inclusion $\mathbb{Q}(\sqrt{2},\sqrt{3}) \leq \mathbb{Q}(\sqrt{2}+\sqrt{3})$. To do this it suffices to show that $\sqrt{2},\sqrt{3} \in \mathbb{Q}(\sqrt{2}+\sqrt{3})$.

 

[Syrus]

So you mean:

Q(√2 + √3) is contained in Q(√2,√3) because √2 and √3 are both in Q(√2,√3), and hence so is √2 + √3 by the field operations.

To see that Q(√2,√3) is contained in Q(√2 + √3) we should show that √2 and √3 are elements in Q(√2 + √3).

This second part seems tricky.. hmm. So far I can show that sqrt(6) is an element of Q(√2 + √3)...

 

[jgens]

It actually is not as tricky as you think. Just consider $\frac{1}{2}[(\sqrt{2}+\sqrt{3})^3-9(\sqrt{2}+\sqrt{3})]$.

Another way to do this problem is to compute the minimal polynomial for $\sqrt{2}+\sqrt{3}$ and see that the degree of the polynomial is $4$. This means that the degree of the extension $[\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}] = 4$. Since $\mathbb{Q}(\sqrt{2}+\sqrt{3}) \leq \mathbb{Q}(\sqrt{2},\sqrt{3})$ and the degree of the extension $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] = 4$, it follows that $\mathbb{Q}(\sqrt{2}+\sqrt{3}) = \mathbb{Q}(\sqrt{2},\sqrt{3})$, as desired. Both techniques have their uses.

 

[Syrus]

Wow, that first part is clever