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# If G is a group- prove that every subgroup of Z(G) is normal in GSolut.docx

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# If G is a group- prove that every subgroup of Z(G) is normal in GSolut.docx

If G is a group, prove that every subgroup of Z(G) is normal in G
Solution
First, the center is a subgroup. We show closure and inverses. Suppose x,y in Z(G). We must show xy in Z(G). Now x,y in Z(G) imply xa=ax and ya=ay for every a in G. Now xa=ax implies x=axa^-1 so that xya = (axa^-1)ya = (axa^1)(ay)=axy. Since now (xy)a=a(xy), by definition xy lies in Z(G). (Here we have used inverses because all of the elements are in G and hence have inverses). Now we show that if x is in Z(G), then x^-1 lies in Z(G). If x is in Z(G) then xa=ax for all a in G. Then xa=ax implies a=x^-1ax which implies ax^-1=x^-1a which means x^-1 is in Z(G). Now that we have showed Z(G) is a subgroup of G, we now show it is normal in G. Let H=Z(G). We must show that H=yHy^-1 for all y in G.Let x be in H. Then we know that yx=xy. But then yxy^-1 = xyy^-1 = x. Hence x is in yHy^-1. On the other hand, suppose x is in yHy^-1. Then x=xyy^-1=yxy^-1 so that xy=yx and hence x is in H. Hence the center is a normal subgroup of G.
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If G is a group, prove that every subgroup of Z(G) is normal in G
Solution
First, the center is a subgroup. We show closure and inverses. Suppose x,y in Z(G). We must show xy in Z(G). Now x,y in Z(G) imply xa=ax and ya=ay for every a in G. Now xa=ax implies x=axa^-1 so that xya = (axa^-1)ya = (axa^1)(ay)=axy. Since now (xy)a=a(xy), by definition xy lies in Z(G). (Here we have used inverses because all of the elements are in G and hence have inverses). Now we show that if x is in Z(G), then x^-1 lies in Z(G). If x is in Z(G) then xa=ax for all a in G. Then xa=ax implies a=x^-1ax which implies ax^-1=x^-1a which means x^-1 is in Z(G). Now that we have showed Z(G) is a subgroup of G, we now show it is normal in G. Let H=Z(G). We must show that H=yHy^-1 for all y in G.Let x be in H. Then we know that yx=xy. But then yxy^-1 = xyy^-1 = x. Hence x is in yHy^-1. On the other hand, suppose x is in yHy^-1. Then x=xyy^-1=yxy^-1 so that xy=yx and hence x is in H. Hence the center is a normal subgroup of G.
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### If G is a group- prove that every subgroup of Z(G) is normal in GSolut.docx

1. 1. If G is a group, prove that every subgroup of Z(G) is normal in G Solution First, the center is a subgroup. We show closure and inverses. Suppose x,y in Z(G). We must show xy in Z(G). Now x,y in Z(G) imply xa=ax and ya=ay for every a in G. Now xa=ax implies x=axa^-1 so that xya = (axa^-1)ya = (axa^1)(ay)=axy. Since now (xy)a=a(xy), by definition xy lies in Z(G). (Here we have used inverses because all of the elements are in G and hence have inverses). Now we show that if x is in Z(G), then x^-1 lies in Z(G). If x is in Z(G) then xa=ax for all a in G. Then xa=ax implies a=x^-1ax which implies ax^-1=x^-1a which means x^-1 is in Z(G). Now that we have showed Z(G) is a subgroup of G, we now show it is normal in G. Let H=Z(G). We must show that H=yHy^-1 for all y in G.Let x be in H. Then we know that yx=xy. But then yxy^-1 = xyy^-1 = x. Hence x is in yHy^-1. On the other hand, suppose x is in yHy^-1. Then x=xyy^-1=yxy^-1 so that xy=yx and hence x is in H. Hence the center is a normal subgroup of G.