Prove that a group G of order 4 is abelian.?

Question

Hints: case I: G is cyclic.
Case II: G is not cyclic- in this case show G can't have an element of order 3; then prove that if all the non-identity elements of a group are of order 2, the group is abelian.

Please try to use these hints to solve.....

Answer 1

Suppose G has an element g of order 3. Then the cyclic subgroup generated be g contains three elements {g, g^2, g^3 = e}, where e is the identity. But the order of every subgroup must divide the order of G, and this is a contradiction. So G has no element of order 3.

G also has no element of order 4, because in Case II, G is not cyclic. Thus G = {e,a,b,c}, where e is the identity, and each of a, b, and c has order 2. Now consider the multiplication table.

__| e | a | b | c |
*****************
e * e | a | b | c |
a * a | e |__|__|
b * b |__| e |__|
c * c |__|__| e |

Now look at the second row. We want to find ab; this cannot be a and cannot be b, so it must be c. Then ac must be b, because every row in the table must conttain evry element in the group.

Similar considerations in the third row tell us that ba = c, and bc = a. You know how to complete the table, and now by inspection, we see G is abelian.

Answer 2

(a) think G is an abelian group with 2 factors, a and b, of order 2. I declare that the subgroup is {0, a , b, ab}. We merely might desire to earnings that the gang satisfies the gang axioms. a million) Associativity - holds directly, because of the fact the operation in our subgroup is comparable to in G. 2) identity - we've 0 because of the fact the identity. 3) Inverses - on condition that a and b are order 2, we've a * a = a^2 = 0 and b * b = 0. subsequently, the inverse of a is a, and the inverse of b is b. Now evaluate the ingredient ab. it is likewise its very own inverse. to work out this: ab * ab = (by utilising commutivity) a^2 * b^2 = 0 * 0 = 0. So, we've shown that {0, a, b, ab} is definitely a set; especially, it particularly is a subgroup of G of order 4. (b) might desire to you make sparkling what you mean by utilising Z(G) and C(a)? all and sundry makes use of diverse notation, so it extremely is not sparkling. i wish this helps!