(Pinter (1982)) AIt's important to note that a group is a set and an operation. If we change one of these, then we either have a different group, or we lose the group classification. Real numbers under addition are a group, as are nonzero real numbers under multiplication, but those are two different groups. Integers under addition are also a group, but a different group than real numbers under addition. Let's prove $\langle\mathbb{B}^{n}, \oplus \rangle$ is a group. Showing a set and operation is a group is pretty algorithmic, in a sense. We just have to show that all three axioms are satisfied.groupis a set $G$ coupled with an operation $\ast$, denoted $\langle G, \ast \rangle$ that satisfies the following three axioms:

(G1: Associativity of the operation)The operation $\ast$ is associative. $a\ast(b\ast c) = (a\ast b)\ast c$ for any $a,b,c \in G$.(G2: Existence of an identity element)There is an identity element inside the set $G$ that we will call $e$ such that for every element $g \in G$, $e\ast g = g\ast e =g$(G3: Existence of an inverse for every element)For every element $g \in G$, there is a corresponding element $ g^{-1} \in G$ such that $g\ast g^{-1} = g^{-1}\ast g = e$

*(G1): Associativity*. This one will be a bit tedious. We have to show associativity for words of any length $n$. Since XOR of words is done bit wise, we can exploit that and first show associativity for binary words of length 1, then "scale it up", if you will. In this case, for words of length 1, we just have to brute-force it. We have to show that for any $a,b,c \in \mathbb{B}^{1}$, that $$(a\oplus b) \oplus c = a \oplus (b \oplus c)$$ So, $$\begin{equation*}\begin{aligned} 1\oplus (1\oplus 1) = 1 \oplus 0 = 1 &\text{ and } (1 \oplus 1) \oplus 1 = 0 \oplus 1 = 1\\ 1\oplus (1 \oplus 0) = 1\oplus 1 = 0 &\text{ and } (1\oplus 1) \oplus 0 = 0 \oplus 0 = 0\\ &\vdots\end{aligned}\end{equation*}$$ Continue in this fashion until you have tried all combinations. Now that we have that this is true for words of length 1, we just use the definition of XOR operation on words of length 1 to "scale up" to words of length $n$. Since the operation is done component-wise, and it is associative on each component, then it is associative on the whole word. We'll show this formally now. Let $\mathbb{a,b,c} \in \mathbb{B}^{n}$. So $\mathbf{a} = a_{1}a_{2}\ldots a_{n}$, $\mathbf{b} = b_{1}b_{2}\ldots b_{n}$, and $\mathbf{c} = c_{1}c_{2}\ldots c_{n}$. Then $$\begin{equation*}\begin{aligned}\mathbf{a}\oplus (\mathbf{b} \oplus \mathbf{c}) &= a_{1}a_{2}\ldots a_{n}\oplus [(b_{1}\oplus c_{1})(b_{2}\oplus c_{2})\ldots (b_{n}\oplus c_{n})]\\ &= (a_{1} \oplus (b_{1} \oplus c_{1}))(a_{2} \oplus (b_{2} \oplus c_{2}))\ldots (a_{n} \oplus (b_{n} \oplus c_{n}))\\&= ((a_{1} \oplus b_{1})\oplus c_{1})((a_{2}\oplus b_{2})\oplus c_{2})\ldots ((a_{n}\oplus b_{n})\oplus c_{n})\\&= (\mathbf{a} \oplus \mathbf{b})\oplus \mathbf{c}\end{aligned}\end{equation*}$$ That third equality holds because we already showed that XOR was bit-wise associative. The last equality just recalls what it means to XOR two binary words. With that, we have shown associativity of the XOR operation.*(G2): Existence of an identity element.*When we want to show that a group has an identity element, we must actually find a candidate and show it meets the criterion. Here is (frustratingly, sometimes), where intuition and experience tend to play the biggest role. My favorite mantra in mathematic is "Make it look like something you've seen before". XOR is bitwise addition, just with a twist (add then mod out by 2). So let's start by considering the identity element for addition: 0. We're looking at binary words of length $n$, so a good candidate for our identity element $e$ would be a string of $n$ 0s. But does it fit? Any word $$\begin{equation*}\begin{aligned}a_{1}a_{2}\ldots a_{n} \oplus 000\ldots 0 &= (a_{1}\oplus 0)(a_{2} \oplus 0)\ldots (a_{n} \oplus 0)\\&= a_{1}a_{2}\ldots a_{n}.\end{aligned}\end{equation*}$$ Check also that $ 0 \oplus \mathbf{a} = \mathbf{a}$, and thus our candidate is a match! Ensure the identity element commutes with any other element. You have to be able to perform the operation on the right and the left, or the candidate fails.*(G3): Existence of an inverse element for every element in the set.*This one is a little bit trickier. We need to be able to find any generic binary word, and show that there is another binary word such that when we XOR them together, we get the sequence of all 0s. (Computer science friends are ahead on this one.) Think back to how we looked at the XOR operation as a form of error checking. If we XORed two words together, there was a 1 in every position in which they differ, and a 0 in every position in which they were identical.Therefore, we come to the interesting conclusion that every element is its own inverse! If you XOR an element with itself, you will get a sequence of 0s. With our error checking interpretation, this makes perfect sense. We know the communication line transmits perfectly if we can XOR the sent and received word and get all 0s every time. Conclusion: every element is its own inverse, and every element in the group is, well, in the group, so we have satisfied the third axiom.