if a group G has three elements show that it is abelian
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Answer:
There is only one group of order 3, the cyclic group of order 3 (which is Abelian). Proof: Let e be the identity element, # the group operation, and g an element of the group other than e. ... Thus g is a generator of the group and has order 3.
When n=1n=1 the group is a trivial one.
Now every group of prime order is cyclic and hence abelian. Hence groups of n=2,3n=2,3 and 55 are abelian.
Since every group of order p2 p2 (where pp is prime) is abelian. Group of order 4= 224= 22 is abelian.✌
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Answer:
G is an abelian group in this case as well.
In either case, we have shown that G is an abelian group.
Step-by-step explanation:
Let G be a group with three elements. Since G is a finite group, every element of G has a finite order. By Lagrange's theorem, the order of each element of G must divide the order of G. Therefore, the possible orders of elements in G are 1, 2, and 3.
If G has an element of order 3, then that element generates G, since G only has three elements. Therefore, G is cyclic, and all cyclic groups of the same order are isomorphic. In particular, all cyclic groups of order 3 are isomorphic to the group Z3, which is an abelian group.
If G does not have an element of order 3, then every non-identity element of G must have order 2. Since every element in G except for the identity has order 2, it follows that G is an abelian group. To see this, let a and b be two elements of G. Then we have:
ab = ba (since a and b commute with each other)
Therefore, G is an abelian group in this case as well.
In either case, we have shown that G is an abelian group.
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