Question

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2. Let G be a group of order 21. Use Lagrange's Theorem or its consequences discussed in class to solve the following problems: (a) List all the possible orders of subgroups of G. (Don't forget the trivial subgroups.) (b) Show that every proper subgroup of G is cyclic. (c) List all the possible orders of elements of G? (Don't forget the identity element.) (d) Assume that G is abelian, so that the map 05: G + G defined by 05(x) = x5 is an endomorphism (as shown on a previous homework). Show that it is in fact an automorphism of G, i.e. show that 05 is one-to-one and onto.

Answer 1

* G is a group of orden 21. By Lagranges thorom, for any finite group G, the order of the every subgroup t of ar divides the order of G. soc Ha packte so the possible orders of s'ubgroups biezties of G are divisars of 94 ie 1,3,7,21 order in the 6 Result: VREG, oca) I IGI ie every element's gurup divides the order of the group 04 H is the proper subgroup of oldes 3 then Hi hao au element of order 3 , sc H i cyclic Similarly If Hy is the proper subgroup of order 7 then its elements has order 1 o77. So 7 an element in H₂ of order 7. Hence He is also cyclic © Eesult: a,bEG & ola) & 0/6 are coprime le . god(ocal, o (6)) = 11 then ocab) = o(a) .0(6). Since, order of the elements of a group divides the order of the group, therefore all possible orders of elements in G are 1,3,7 & if at with o(a)=3 & beG with o (6)=7 Then they siuee Coca), 0(6)) = (,7) - 1 therefore o(ab) = 3.7 = 21 Some Goto Hasean element So, 24 is also a possible order of an element of a (Then G became cyclic)

@ @ kel G is abelian ds: G G is defined by (x) = 205 is an endomorphism. e. We have to show do is one-one &e onto one-one : Let xe keras then do(a) = e (e is the identity element) ie x5=e → x= e siuce G has no element of order 5) This implies kerds = {e}, Therefore, me-one. Onto: Since of is one-one & Gis of finite order, it & also onto.