Essay: The Brain in a Vat Argument

There are many different kinds of stances that philosophers can take, ranging as far as the many topics that are chosen to discuss and focus upon. The philosophy of skepticism has existed in varying forms since ancient Greece such as philosopher Gorgias the Nihilist who stated that nothing exists, even if something existed there was no way to know, even if we could know that it would be impossible to communicate it. In 1986, American philosopher John Pollock published a short two page story in “Contemporary Theories of Knowledge” introducing the philosophical topic of skepticism through the concept of being a brain in a vat. In this story, Mike is living a normal life until a man named Harry is taken by suspicious characters whereupon Mike discovers that there are a group of people transferring people’s brains into vats for science and that Mike was transferred months prior. Mike concludes the story by pondering the concept of reality.

“I am racked by the suspicion that I am really a brain in a vat and all this I see around me is just a figment of the computer. After all, how could I tell? If the computer program really works, no matter what I do, everything will seem normal. Maybe nothing I see is real. It’s driving me crazy. I’ve even considered checking into that clinic voluntarily and asking them to remove my brain just so that I can be sure.” (p.195)

Skeptics employ arguments such as this one to state that nothing can be known for certain because there is no way to test the reality that is perceived without relying upon the senses that said reality provides. Pollock is arguing here that there is no way to verify if reality is all a lie because he happens to be a brain in a vat or if reality is true so no conclusions can be drawn about anything; nothing can be truly known.

There is a lot to be said for the brain in a vat argument; mainly that it takes away the burden of proof. If you cannot trust or allow evidence of the physical world to be used as valid points of argumentation then the philosophical inquiry becomes ‘prove me wrong’ rather than ‘let me show you why I’m right’. This allows Pollock to operate from a stance of superiority because the argument redefines the rules: the senses deceive so any evidence to the contrary that relies on the senses will be ignored. In the argument Pollock presents in the story,

“The computer monitors the output of his motor cortex and provides input to the sensory cortex in such a way that everything appears perfectly normal to Harry. It produces a fictitious mental life so that he is unaware that anything has happened to him. He thinks he is shaving right now and getting ready to go to the office and stick it to another neurosurgeon. But actually, he’s just a brain in a vat.” (p.195)

Pollock outlines the extent to which disbelief can be wrought. Effectively, since so much sensory output comes from the brain, if the brain is deceived then all sensory data can be overwritten falsely. In a similar fashion, Descartes stated cogito ergo sum to conclude that from a skeptic’s perspective, the only information that can be rationalized to is ones own existence by process of being a thinking thing which for Pollock meant existing as a brain in a vat.

That said, the brain in a vat argument has some steep limitations to its realism. During encounters with other individuals, one can learn and process new information, expand upon previous knowledge, and discuss a multitude of viewpoints that would not have been considered prior to the conversation. Carefully detailed surprises and experiments that give unexpected or unknown results are examples of information in the world that is outside of the self. The type of mechanism required to create enough sensory data to operate the brain in a vat such that the brain itself would not be able to notice the inconsistencies would require a matrix system larger than the ones used to encrypt the internet. The efforts that would be necessary to make this happen are exorbitant, raising the question: why was the brain not given sensory data in order to stimulate a depressive state so that the details would not be as relevant or noticed; creating a system that could afford to be a lot smaller and have an insignificant realm of influence? While the senses may deceive so that everything experienced could indeed be a falsification, logic remains as true thus consequences that are deemed unsupported and unlikely do not necessarily need to be proven wrong to be dismissed. While the brain in a vat scenario attempts to put the burden of proof on falsifying the claim, reasoning through complex examples and the structure therein that would require such a scenario to be likely scopes enough of the constraints to argue its opposition. In the passage, Pollock writes,

“I’ll bet you think we’re going to operate on you and remove your brain just like we removed Harry’s, don’t you? But you have nothing to worry about. We’re not going to remove your brain. We already did—three months ago!” (p.195)

If the brain in a vat scenario were likely, then the mechanism keeping the brain falsely stimulated should have enough self preservation to make it impossible for the brain to question its reality in a vat; in the story the protagonist is told of the situation and sees it unfold firsthand which would be counter productive and harmful to the science experiment by allowing the thought to occur because the secret of the reality could be unearthed.

The concept of existing as a brain in a vat is often gone to for an example of a skepticism because it demonstrates a simplistic scenario presenting a reason to doubt what is seen and perceived as reality. That being stated, if it impossible to distinguish the difference between reality and being a brain in a vat, then the argument holds no weight because there is no significance to outcome; life will be lived as perceived, the brain cannot be released by ones own devices, the question under consideration cannot be tested nor verified. As such, this argument is not a good example for skepticism.

BIBLIOGRAPHY

Feinberg, Joel and Shafer-Landau, Russ. “Reason and Responsibility.” Cengage Learning 2015: pages 193-195. Print.

Poetry: Wanting Release

My head hurts
My heart hurts

I want to close my eyes
I wish screaming took pain away

Take all the stress that has interwoven itself into my skin
Yank it out
Turn it into a ball
(Ironic, right?)

Burn it away
What is left of me?
Doesn't matter

I need release.

Poetry: Self Love

Sometimes it takes someone
Looking at my flaws
The parts I cover up
Exposing them to the light
No judgement in blank curiosity
Pure desire to know me, see me
Discover where I'm at
Spend a moment embracing it
Embracing me
Full clarity
I'm not hiding me

How you look at me then
Placing value and worth
Upon my description...
I know then that I can too

So I thank you
Thank you for your kindness
Thank you for wanting what I have to offer
Thank you for spending this moment together

I can learn to love me too.

Poetry: Forward

Moving on is like unmasking an onion

What is its center?

Does anyone really know?

Peel back layer by layer

Forget what your life has done to you

Freshen up

Free yourself

Forgive yourself

What others view is not what you are

Allow it to go

Poetry: Relieved

The nightmare is over
I have woken up
The wounds inflicted
Are naught but a memory

The people are my demons
No longer visible
I am alone

The emotional trauma
A daily war

Over

Poetry: It's Just How I Feel

Some days I don't feel loved at all by anyone in any amount

Some days I don't feel needed and when I try, I become a bother

Some days I don't feel worthy so if I'm praised, ignored, or yelled at I interpret it all the same

Some days I don't know how to get out of bed

Some days I don't know how to eat

Some days I don't know how to carry on

Some days it's more often than that.

Poetry: I Loved

You can't make me hate you
You can only make me hate myself

Essay: An Introduction to Sylow's Theorems

1 Introduction
In 1872 while studying Galois Theory, Ludwig Sylow wrote a paper that contained some theorems that are now known as Sylow’s Theorems. The content of these theorems focuses on how many subgroups of fixed order a given finite group holds [18, p. 280]. Why are these theorems important? For starters, given the proof of Sylow’s we can establish Cauchy’s theorem via induction. On the reverse, Cauchy’s Theorem is often used as a method to prove Sylow’s Theorems [18, p. 281]. In addition to this, these theorems play a large role in finite group theory as they aid greatly in classifying finite, simple groups.

For the purposes of this paper, let p be a prime number, G be a finite group, e be the identity of G, Z(G) be the center of G, and |G| be the order of group G.

2 Definitions 

2.1 Sylow p-Group
A group is a Sylow p-group if the order of the group is a power of p.

2.2 Sylow p-Subgroup
If pk is the highest power of a prime p dividing the order of finite group G, then a subgroup of G of order pk is called a Sylow p-subgroup of G.

Note
Sometimes this is referred to as a p-Sylow subgroup of G. Example
Let us look at D6. The elements of D6 are {e, r, r2, r3, r4, r5, s, rs, r2s, r3s, r4s, r5s}. The only 3-Sylow subgroup is the set {e, r2, r4} but there are three 2-Sylow sub- groups: {e, r3, s, r3s}, {e, r3, rs, r4s}, {e, r3, r2s, r5s}.

2.3 np (G)
Let the number of Sylow p-subgroups of G be denoted np(G) = |Sylp(G)|.

3 Lagrange’s Theorem
The order of any subgroup of a group must divide the order of the group.

3.1 Proof
Let H be a subgroup of G. We want to show that |H| divides |G|. We know that the right cosets of H form a partition in G so every element of G belongs to at least one right coset of H in G. We also know that there is no element that belongs to two distinct right cosets of H in G. Thus, every element of G belongs to exactly one right coset of H. We know that every right coset of H contains |H| elements so |G| = u|H| with u being the number of right cosets of H in G. Therefore |H|||G|.
QED

Note
The converse is not true. For example, A4 = {1, 2, 3, 4} has order 12 but there does not exist a subgroup of order 6. Sylow’s Theorems provides more information as to what the properties of subgroups of a certain group are which is why Sylow’s Theorems are sometimes considered a partial converse to Lagrange [8, p. 1].

4 Cauchy’s Theorem
Let p divide |G|. Then G has an element of order p.

4.1 Proof
Let S = {a1,a2,a3,...,ap} such that every ai is in G and a1a2a3 ...ap = e for |G| = n. We know S has np−1 elements. Let us define a relation on S where two sets with p elements are equivalent if one set is a cyclic permutation of the other set.
Case 1
If all components of a set with p elements are equal then its equivalence class has one member.
Case 2
If two components of a set with p elements are distinct then the equivalence class has p members.
Let the number of solutions to xp = e be called r. Then r is the number of equivalence classes with just one member. Let s denote the number of equiv- alence class with p members so np−1 = r + sp thus p|r.
QED

Note
The proof for Cauchy’s Theorem and Sylow’s 1st Theorem are very similar in formatting. Oftentimes, Cauchy’s Theorem is considered to be a special case of Sylow’s 1st Theorem [2, p. 1]. While originally Cauchy was used for a finite, abelian group, it has since been shown that Cauchy holds for any finite group regardless of whether or not it is abelian [16, p. 3].

5 Sylow’s 1st Theorem
If pk divides |G| then G has a Sylow p-subgroup.

5.1 Proof
Let pk divide |G|. We want to show G has a subgroup of order pk.
Case 1:
Suppose p divides the order of Z(G). By Cauchy’s Theorem for abelian groups, Z(G) must have an element of order p, denote it m. By induction, the quotient group G/ < m > for abelian G has a subgroup R of order pk−1 so the preimage of R in Z(G) is a subgroup of order pk.
Case 2:
Suppose p does not divide the order of Z(G). Let C1, C2, . . . , Cn be the conju- gacy classes in G with more than one element. Then |G| = |Z(G)| + |C1| + · · · + |Cn|.
Since p by definition does not divide the order of the center, there exists at least one conjugacy class whose order is not divisible by p. Let us denote this conjugacy class as cong(c). We know that
|cong(c)| = [G : CG(c)] = |G|/|CG(c)|.
Note that pk must divide the order of subgroup CG(c) so by induction, G has a Sylow p-subgroup.
QED

6 Sylow’s 2nd Theorem
In a finite group with p dividing |G|, all the Sylow p-subgroups are conjugate for some fixed p.

6.1 Proof
View [1, 2, 9, 12] to examine how this proof can be done using orbits.
QED

7 Sylow’s 3rd Theorem
The number of Sylow p-subgroups for a fixed p is congruent to 1 ≡ mod p and divides |G|.

7.1 Proof
Let P be a Sylow p-subgroup that acts on the set S = {P = P1, P2, ..., Pr} via conjugation. We know from Sylow’s 2nd Theorem that the only P conjugate of P is itself and the the other conjugacy classes have orders of some power of p. This will be shown in more detail in the note below. Let |S| be the sum of positive powers of p so |S| ≡ 1 mod p. Suppose G acts on |S| via conjugation. Since all p-Sylow subgroups are conjugate, there exists only one orbit so for P in S, |S| = |P| = [G : N(P)]. However, [G : N(P)] must divide |G| so np(G) must divide |G|.
QED

7.2 Note
There are a multiple of ways to solve Sylow’s second and third theorems includ- ing separating the statements into parts and solving them in varying orders. To show an example of this, let us look at part of theorem three:
The number of Sylow p-groups for a fixed p is congruent to 1 ≡ mod p.

Proof
We want to show that np(G) ≡ 1 mod p. Let H be a Sylow p-subgroup that acts on the set S = {H = H1,H2,...,Hr} via conjugation where all Hi are Sylow p-subgroups. We know G acts on S by conjugation. Since H < G, H also acts on S by conjugation.
Consider the orbits of H as shown in [1, 2, 9]. The order of an orbit is the index (P : Ps) where (H : Hs)||H|. Since the order of H is a power of p, the order of each orbit is also a power of p. Then the order of an orbit is divisible by p except in the case p0 =1 where hHih−1 =Hi for some i and all h∈H so h ∈ Hi. Since Hi is a Sylow p-subgroup, H = Hi. This means that if an orbit consists of a single element, it must be {H}. Conversely, for h ∈ H, hHh−1 = H. Then the orders of the orbits of S under conjugation by H are all divisible by p except for {H}. It follows that |S| is congruent to 1 mod p where |S| is the number of conjugates of H.
QED

8 Example
Let G be a group of order pq with p<q where p and q are both prime. Then G has exactly one subgroup of order q, which is a normal subgroup of G.

8.1 Proof

We know nq has to divide pq/q=p so nq =1 or p. We also know nq ≡1mod q. Since p>1 and p<q then p<q+1 so nq =1. Thus the Sylow q-subgroup with order q is the only one, making it a normal subgroup of G.
QED

9 Corollary
Let P be a Sylow p-subgroup of G. The following are equivalent.
P is the unique Sylow p-subgroup of G so np(G) = 1.
P is normal in G.
All subgroups generated by elements of the p-power order are p-groups. In other words, if X is a subset of G such that |x| is a power of p for all x in X then < X > is a p-group.

10 Theorem
The groups A5 and S5 each have ten subgroups of size 3 and six subgroups of size 5.

10.1 Proof
Any element with odd order in a symmetric group is an even permutation thus subgroups 3-Sylow and 5-Sylow are in A5. We know |A5| = 60 = 22 ∗3∗5 so the 3-Sylow subgroups have size 3 and the 5-Sylow subgroups have size 5. Let us denote these numbers n3 and n5 respectively. By Sylow’s 3rd Theorem, n3|20 and n3 ≡ 1 mod 3 thus n3 = 1,4, or 10. There are twenty 3-cycles in A5 so there are ten subgroups of size 3 since the 3-cyles come in inverse pairs, making n3 =10. For the 5-Sylow subgroup, n5|12 and n5 ≡1mod5 thus n5 =1, or 6. There are at least two subgroups of size 5 in A5 so n5 >1 and n5 = 6.
QED

11 Normal Sylow Subgroups Theorem
The condition np = 1 means that a p-Sylow subgroup is a normal subgroup.

11.1 Proof
We know by Sylow’s 2nd Theorem that all p-Sylow subgroups are conjugate. Then np = 1 when a p-Sylow subgroup of G is self conjugate thus a normal subgroup of G.
QED

12 Theorem

If p and q are different prime factors of |G|, np = 1, nq = 1 then the elements of the p-Sylow subgroup commute with the elements of the q-Sylow subgroup.

12.1 Proof
Let P be the p-Sylow subgroup, Q be the q-Sylow subgroup. Since P and Q have relatively prime sizes, P∩Q = {e} by Lagrange’s Theorem. We know that the subgroups P and Q are normal in G so for a in P and b in Q, aba−1 = (aba−1)b−1 = a(ba−1b−1) in P∩Q = {e} so ab = ba.
QED

13 Example
Prove that if |G| = 30 then the group is not simple.

13.1 Proof
Assume |G| = 30. We know 30 = 2∗3∗5 and that there exists a number of Sylow p-subgroups for each p such that p||G|. For p = 2, n2|(3 ∗ 5) and n2 ≡ 1 mod 2. The factors of p = 2 would then be 1, 3, 5, 15 with all factors congruent. For p=3, n3|(2∗5) and n3 ≡1mod3. The factors of p = 3 would then be 1, 2, 5, 10 with 1, 10 congruent. For p = 5, n5|(2 ∗ 3) and n5 ≡ 1 mod 5. The factors of p = 5 would then be 1, 2, 3, 6 with 1, 6 congruent.
Since there are several numbers that could possibly work for the primes, suppose to show a contradiction that n3 = 10 and n5 = 6. Then n3(G) = 10 so |n3(G)| = 3 elements, two of which are non identity, giving a total of 21 elements. Repeat for n5 = 6 to get 24 elements total. We know 21+24 = 45 > |G| = 30, which is a contradiction. Therefore, n3 = 1 or n5 = 1 so n3(G) or n5(G) is characteristically normal in G thus G is not simple.
QED

14 Conclusion
Sylow’s Theorems may appear to be a simple list on finite group properties but the applications of Sylow’s Theorems are rather diverse including classifying groups with order pq with p < q for p and q both prime, determining if a finite group is nilpotent, determining normality of a Sylow subgroup, determining if a group of a certain size is abelian, determining if a group of a certain order is simple, and so much more.

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