cyclic groups help **Question 12:** Prove that the following groups are *not* cyclic:(a) \(\mathbb{Z}_2 \times \mathbb{Z}_2\)(b) \(\mathbb{Z}_2 \times \mathbb{Z}\)(c) \(\mathbb{Z} \times \mathbb{Z}\)

To show the following groups are not cyclic. If a group is cyclic the there exit an element in that group which generates the whole group. (a) Z2×Z2 To show that Z2×Z2 is not cyclic. The group Z2×Z2=0,0,1,0,0,1,1,1. To show Z2×Z2 is not cyclic first we have to find the order of each non-zero elements. 21,0=0,020,1=0,021,1=0,0 So we can conclude that every element has at most order 2, so no element has order 4, thus there is no element generates the whole group(Since the order of generator must be 4). Therefore the group Z2×Z2 is not cyclic.(b) T o show that Z2×Z is not cyclic: Suppose the contradiction i.e. Z2×Z is a cyclic group. Then there is an element x∈Z2×Z with,Z2×Z=x. So x=(n,m) for some integer m∈Z & n∈Z2. So, Z2×Z=x=xk: k∈Z=kn,km: k∈Z. If m=0 then (0,1) is not in the set, which is a contradiction. So we have m≠0. If n=0 then (1,0) is not in this set, which is a contradiction. So we have n≠0. Consider the element 1,2m∈Z2×Z. Then, there is an integer k∈Z with kn,km=1,2m, and since n,m≠0 this gives , k is odd (Since kn=1& n∈Z2, n≠0 thus n=1) Also km=2m⇒k=2. This gives k is odd and k=2⇒a contradiction. So, Z2×Z is not cyclic. To show that Z×Z is not cyclic: Assume the converse is true i.e. Z×Z is a cyclic group. Then there is an element x∈Z×Z with Z×Z=x. So, x=(n,m) for some integer m∈Z & n∈Z. So, Z×Z=x=xk: k∈Z=kn,km: k∈Z. If m=0 then (0,1) is not in the set, which is a contradiction. So m≠0. If n=0 then (1,0) is not in the set, which is a contradiction. So n≠0. Consider the element n,-m∈Z×Z. There is an integer k∈Z with kn,km=n,-m, and since n,m≠0 this gives k=-1 and k=1, which is a contradiction. S o Z×Z is not cyclic.