# 2.The Orwell Group

## 08.06.2020

### Contents/Index

@2. The Orwell Group

I have long used the notion of $2 + 2 = 5$ to describe serious absurdities. This is taken from 1984 by Orwell. The book centers around two different themes, both about how to properly control a society. The one is total surveillance. The second is through language. Here we sort of have two ways to go about. Both are based on the idea that we perceive and understand the world around us through our language. The one way is to reduce the expressiveness of the language. That is remove words from it. The second is to constantly alter the meaning of the language. The fact that $2 + 2 = 4$ is central in our way of reasoning. I can figure this out by counting my fingers or rocks or whatever. So a language in which $2 + 2 = 5$ doesn't seem very sound.

Anyway we can construct a model for group theory (or construct a group, for short) in which $2 + 2 = 5$. We call this group The Orwell Group. Let's go!

Let $\mathcal{M}_{TOG} = (M,+)$ where $$M = \{0,2,5\}$$ and where $+$ is given as

• $0 + a = a + 0 = a$ for all $a$ in $\mathcal{M}_{TOG}$.
• $2 + 5 = 5 + 2 = 0$
• $5 + 5 = 2$
• $2 + 2 = 5$

### Proof of Group

So we clearly have an identity element $0$. We have an additive inverse for each element. The group is abelian, that is we have $$a + b = b + a, \forall a,b \in \mathcal{M}_{TOG}$$ This can be seen directly from how $+$ is given.

The group is associative. That is $$(a + b) + c = a + (b + c)$$ I think there's two ways of proving this. The one I'm a bit unsure of. But it seems that we should be able to put parentheses in a way that doesn't matter, but according to how we read an expression. Eg. from left to right: $$a + b + c = (a + b) + c$$ Since abelian we have that $$a + b + c = b + c + a$$ Again we have that $$b + c + a = (b + c) + a = a + (b + c)$$ So in total we have that $$(a + b) + c = a + (b + c)$$ Another way to prove it is sort of the hard way - exhaust possibilities. However if $$a = 0 \lor b = 0 \lor c = 0$$ we are done proving. The same goes for the situation where $$a = b = c$$ In the last situation we must have $$a = b \neq c \lor b = c \neq a \lor a = c \neq b$$ Where $a,b,c \in \{2,5\}$. For $a = b \neq c$ we get $$(a + b) + c = c + c = a \land a + (b + c) = a + 0 = a$$ The same thing happens for $b = c \neq a$. For $a = c \neq b$ we get $$(a + b) + c = 0 + c = c = a \land a + (b + c) = a + 0 = a$$ Done. Or we could skip all this and observe that $\mathcal{M}_{TOG} \cong \mathbb{Z} / 3 \mathbb{Z}$.

### Conclusion

One key point here is that we are extremely liberal on how to interpret terms. We might as well replace $5$ with $a$ and $2$ with $b$ and still have the same structure. Another point is that we have just defined $+$ as we see fit. Yet another point is regarding the "idea". What IS the POINT of $\mathcal{M}_{TOG}$?