Question

Q. prove that a non-commutative group has at least six elements.

Answer 1

Answer on Question #59928 – Math – Abstract Algebra

Question

Prove that a non-commutative group has at least six elements.

Proof

Lagrange's theorem shows that any group of prime order is cyclic and therefore commutative. It is obvious that

1. If group has 1 element, it is commutative: $\mathbb{G} = \{e\} : ee = e$.

2. If group has 2 elements, it is cyclic of order 2 and commutative:

$\mathbb{G} = \{e, a\}$: $a^2 = e$, $a = a^{-1}$, $aa^{-1} = e = a^{-1}a$.

3. If group has 3 elements it is cyclic of order 3 and commutative:

$\mathbb{G} = \{e, a, b\}$: $a^3 = e$, $b = a^{-1} = a^2$, $ab = ba = e$.

4. If group has 4 elements, it is isomorphic either to

a) cyclic group of order 4: $\mathbb{G} = \{e, a, b, c\}$: $a^4 = e$, $b = a^2$, $c = a^3$, $a^3 = a^{-1}$, $ab = ba = c$, $ac = ca = e$, $bc = cb = a$;

or to

b) Klein four-group: $\mathbb{G} = \{e, a, b, c\}$: $a^2 = b^2 = c^2 = e$, $ab = ba = c$, $ac = ca = b$, $bc = cb = a$.

5. If group has 5 elements, it is isomorphic to the cyclic group of order 5, which is commutative:

$\mathbb{G} = \{e, a, a^2, a^3, a^4\}$.

Thus, if group has 5 or less elements, there is no possibility to define a non-commutative group operation.

If there are 6 or more elements, we have "enough space" to introduce a non-commutative group operation. The symmetric group $S_3$ has 6 elements and it is non-commutative, because commutativity does not hold for all elements of $S_3$.

For example,

(1 2)(1 3) = (1 3 2), (1 3)(1 2) = (1 2 3), that is, (1 2)(1 3) ≠ (1 3)(1 2) (non-commutative elements);

though (1 2 3)(1 3 2) = e, (1 3 2)(1 2 3) = e, that is, (1 2 3)(1 3 2) = (1 3 2)(1 2 3) (commutative elements).

Answer: we have proved that group has to have at least 6 elements to be a non-commutative one.

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