1. prove no two of C_8, C_4 X C_2, C_2 X C_2 X C_2 are isomorphic.C_n is cyclic group of order n.
Give, with justification, a group of order 8 that is not isomorphic to any of those groups.
2. Prove that if m and n are coprime then C_m X C_n is cyclic.
if m and n are not coprime. can it be cyclic?
3. H is a normal subgroup of a finite group G. Which of the followings are true?
(i) if G is cyclic then H and G/H are cyclic.
(ii) If H and G/H are cyclic then G is cyclic.
(iii) If G is abelian then H and G/H are abelian
(iv) If H and G/H are abelian then G is abelian.
4. find all homomorphisms between C_11 to C_14
5. H is a subgroup of G and H not equal to G. Show that there is an element of G which does not belong to any subgroup of the form gHg^(-1) for g belongs to G
is cyclic because every subgroup of a cyclic group is cyclic. Furthermore, is cyclic because there exists such that this means generates .
False. Let and its 3-element subgroup.(ii) If H and G/H are cyclic then G is cyclic.
Yes. Define the natural map then is a homomorphic image of an abelian group so it must be abelian. Or you can say since .(iii) If G is abelian then H and G/H are abelian
False. Use the same example.(iv) If H and G/H are abelian then G is abelian.
Maybe I am missing something really easy here. But I can only prove it for finite groups , I hope that is what is being ask for. Say are subgroups such that is a subgroup. Consider where because it is closed. So then without lose of generality. So and so on until , permuting these products it means . Thus, it means . Now assume is finite. If every is contained in the conjugate subgroup for some then it means this union is finite for we assumed was finite. But by above this union is simply for some . But that means and since means . We have shown that if has this property (for a finite group) then by above argument.
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Now it would be nice if this union can generalize, i.e. if is any group then for implies for some . But sadly this does not work. Consider to be an -tuple . Now let . Then they unionize to . But no one groups contains them all.