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December 19th, 2018, 10:27 AM  #1 
Newbie Joined: Dec 2018 From: Darjeeling, India Posts: 4 Thanks: 1 
Let G be a group in which (ab)^2 = (a^2)(b^2) for all a,b ∈ G. Show that H = { g^2g ∈ G } is a normal subgroup of G. Last edited by skipjack; December 19th, 2018 at 11:36 AM. 
December 19th, 2018, 12:02 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,105 Thanks: 1907 
Hint: the condition given for G is equivalent to commutativity.

December 19th, 2018, 06:33 PM  #3 
Newbie Joined: Dec 2018 From: Darjeeling, India Posts: 4 Thanks: 1 
Can you show me how please.

December 19th, 2018, 06:57 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,105 Thanks: 1907 
(ab)² = (a²)(b²) can be written as a(ba)b = a(ab)b, which is equivalent to ba = ab. It's now easy to show that H is a subgroup of G, so it's a normal subgroup. 
December 19th, 2018, 07:01 PM  #5 
Newbie Joined: Dec 2018 From: Darjeeling, India Posts: 4 Thanks: 1 
Dear Sir, I see that our motive is to show that G is abelian. But I think that in this case we need to show that: a*b=b*a i.e., (ab)²=(ba)² i.e., a²b²=b²a². Can you enlighten me regarding this. And the question holds only 2 marks. So kindly give me outline of steps of my answer as I don't think it's practical to show that H is a subgroup first & then normal using the axioms & definitions for 2 marks. 
December 19th, 2018, 07:27 PM  #6  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,980 Thanks: 789 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
December 19th, 2018, 07:57 PM  #7 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 287 Thanks: 88 Math Focus: Number Theory, Algebraic Geometry  
December 19th, 2018, 08:31 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,105 Thanks: 1907 
As the group operation is associative, (ab)² = (a²)(b²) is equivalent to a(ba)b = a(ab)b. a(ba)b = a(ab)b implies a$^{1}$a(ba)bb$^{1}$ = a$^{1}$a(ab)bb$^{1}$, which simplifies to ba = ab. Conversely, ba = ab implies a(ba)b = a(ab)b. 
January 5th, 2019, 01:09 PM  #9  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,641 Thanks: 119  Quote:
H is normal sub group if gH=Hg for all g in G where gH={ghh$\displaystyle \epsilon$H} Hg={hgh$\displaystyle \epsilon$H gh=gg$\displaystyle ^{2}$=g$\displaystyle ^{3}$=hg regardless of OP conditions Hopefully I survive this post. After the last one my screen went blank, and I notice a red warning sign next to home button "Not secure" Last edited by zylo; January 5th, 2019 at 01:19 PM.  
January 5th, 2019, 02:11 PM  #10  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,980 Thanks: 789 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  

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