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December 19th, 2018, 10:27 AM   #1
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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 11:36 AM.
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December 19th, 2018, 12:02 PM   #2
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Hint: the condition given for G is equivalent to commutativity.
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December 19th, 2018, 06:33 PM   #3
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Can you show me how please.
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December 19th, 2018, 06:57 PM   #4
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(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.
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December 19th, 2018, 07:01 PM   #5
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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.
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December 19th, 2018, 07:27 PM   #6
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Quote:
Originally Posted by Prasant Thakuri View Post
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
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December 19th, 2018, 07:57 PM   #7
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Quote:
Originally Posted by Prasant Thakuri View Post
(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.
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December 19th, 2018, 08:31 PM   #8
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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.
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January 5th, 2019, 01:09 PM   #9
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Quote:
Originally Posted by Prasant Thakuri View Post
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"
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Last edited by zylo; January 5th, 2019 at 01:19 PM.
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January 5th, 2019, 02:11 PM   #10
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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.)

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