Solution: Let H be a subgroup of an abelian group G. If x belongs to H and g belongs to G, then gxg^{-1} = gg^{-1}x = x belongs to H.
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2.
Let G_{1} and G_{2} be groups,
and let G be the direct product G_{1} × G_{2}.
Let H = { (x_{1},x_{2}) in G_{1} × G_{2}
| x_{2} = e } and
K = { (x_{1},x_{2}) in G_{1} × G_{2}
| x_{1} = e }.
Prove that H and K are normal subgroups of G.
Note: Exercise 3.3.9 in the text
shows that H and K are subgroups
for which HK = G and
HK = {e}.
Solution: If (x_{1},e) is any element of H, and (g_{1},g_{2}) is any element of G_{1} × G_{2}, then
(g_{1},g_{2})
(x_{1},e)
(g_{1},g_{2})^{ -1}
=
(g_{1},g_{2})
(x_{1},e)
(g_{1}^{ -1},g_{2}^{ -1})
=
(g_{1}x_{1}g_{1}^{ -1},
g_{2} e g_{2}^{ -1})
=
(g_{1}x_{1}g_{1}^{ -1},e).
Since (g_{1}x_{1}g_{1}^{ -1},e) belongs to H, this shows that H is a normal subgroup of G. A similar argument shows that K is also a normal subgroup of G.
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