My Math Forum problem on normal subgroups

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 December 19th, 2018, 09:27 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^2|g ∈ G } is a normal subgroup of G. Last edited by skipjack; December 19th, 2018 at 10:36 AM.
 December 19th, 2018, 11:02 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,628 Thanks: 2077 Hint: the condition given for G is equivalent to commutativity.
 December 19th, 2018, 05:33 PM #3 Newbie   Joined: Dec 2018 From: Darjeeling, India Posts: 4 Thanks: 1 Can you show me how please.
 December 19th, 2018, 05:57 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,628 Thanks: 2077 (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, 06:01 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, 06:27 PM   #6
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Quote:
 Originally Posted by Prasant Thakuri 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.
Does it really matter if the problem is "only" worth 2 marks? If you are taking the class then you should be able to do this problem, else it wouldn't have been assigned. And with knowing how to solve this you might be better prepared to do a question worth more points.

-Dan

December 19th, 2018, 06:57 PM   #7
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Quote:
 Originally Posted by Prasant Thakuri (ab)²=(ba)² i.e., a²b²=b²a²
The statements here are equivalent, but not (I suspect) for the reason you think. $(ab)^2 = a^2b^2$ it is not true in general - it's true if and only if $a$ and $b$ commute.

 December 19th, 2018, 07:31 PM #8 Global Moderator   Joined: Dec 2006 Posts: 20,628 Thanks: 2077 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. Thanks from topsquark
January 5th, 2019, 12:09 PM   #9
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Quote:
 Originally Posted by Prasant Thakuri Let G be a group in which (ab)^2 = (a^2)(b^2) for all a,b ∈ G. Show that H = { g^2|g ∈ G } is a normal subgroup of G.
Amazes me that no one begins with a definition:

H is normal sub group if gH=Hg for all g in G where
gH={gh|h$\displaystyle \epsilon$H}
Hg={hg|h$\displaystyle \epsilon$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 12:19 PM.

January 5th, 2019, 01:11 PM   #10
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 Originally Posted by zylo 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"
Yeah, I've been seeing that a lot, too. I have no idea why the different sites are doing this. (I'm getting messages for three of them.)

-Dan

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