11. Let G=  2   8 ,
H={(0,0), (1,0)}, K={(0,0), (0,1), (0,2), (0,3), (0,4), (0,5), (0,6), (0,7)},
S={(0,0), (1,4)}, T={(0,0), (1,1), (0,2), (1,3), (0,4), (1,5), (0,6), (1,7)}.
It is easy to check that, H, K, S and T are subgroups of G, also
H  K=G and S  T=G, that is each of these subgroups is direct summand, hence
they are pure subgroups. On the other hand:
H+S= {(0,0), (1,0), (1,4), (0,4)} is not a direct summand (check!) hence not pure
( by Theorem7). And K  T={(0,0), (0,2), (0,4), (0,6)} which is not a direct
summand too (check!), hence not pure for the same reason.
12. Assume that G is a group such that each of its subgroups is pure, then a  G
implies a is pure. If a is not torsion, and 0  n   , then na is a pure
subgroup of G too. By purity of na we have na= n(mna),for some integer m so
n(1-mn)a=0, a contradiction to the assumption that a is not torsion. Now assume
that p 2 o(a ) . By hypothesis, the subgroup p a is pure in G, which implies
p a  p (mp a ) , for some integer m, hence p (1  mp )a  0 . Assume o(a)=k ,
we have : k =s p 2 and p (1  mp )  tk , by substitution we get p(1-mp)=tsp 2
, hence (m+ ts )p=1 , not possible. Therefore, o(a) is square free.
I leave the converse for the students.
13. (a) Assume that T is a pure subgroup of G, and S is a subgroup of T. Let
x  T and x+ S= n(g+ S), where n   and g  G , then x-ng  S, say x-ng=s,
where s  S, then x-s=ng, and x-s  T ( since S  T). T is pure implies x-s =nt for
some t in T then x-nt=s  S, which implies x+S=n(t+S) , where t+S  T S . This
proves that T S is pure in G S .
(Note that it was not needed the condition S is pure).
(b) The same example of ex. 11 , H and S are pure in G but (S  H) H is not
pure in G H (Check!).