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My question is this. I have Let $H\le G, a,b\in G$ Define $Hx=\{hx|h\in H\}$ Show that $Ha=Hb$ or $Ha\cap Hb=\emptyset$.

I thought I would do a proof by contradiction. So suppose that $Ha\cap Hb$ is nonempty. Then there exists an element $x$ such that $x\in Ha$ and$ x\in Hb$ Thus we can write $x$ as $$x=h_1a,\text{ and } x=h_2b$$ Then $h_1a=h_2b$ and thus $a=h_1^{-1}h_2b$.

This is where I'm getting stuck. I just don't know how to go about showing that there exists a contradiction.

Greg Martin
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Iceman
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  • Show that $a\sim b \iff ab^{-1}\in H$ is an equivalence relation, and $Hb$ are the equivalence classes. – Pedro May 17 '14 at 03:30
  • I like that method, but I'm trying to follow the book I'm working out of, and since equivalence relations are not covered until the next section, I want to use what I have done up to now...basically subgroups is the section the exercise is under... – Iceman May 17 '14 at 03:33
  • Equivalence relations are very basic mathematicals objects. You might as well get used to using them. – Pedro May 17 '14 at 03:34
  • there is an answer here... http://math.stackexchange.com/questions/737916/let-h-be-a-subgroup-of-g-and-let-a-and-b-belong-to-g-then-ah-bh-or?rq=1 – Eleven-Eleven May 17 '14 at 03:35
  • What book are you using? – Lemon May 17 '14 at 03:37
  • Abstract Algebra by Herstein – Iceman May 17 '14 at 03:39
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    Perhaps your problem is because there isn't a contradiction to be had! It is very possible for $Ha\cap Hb$ to be nonempty. You're not trying to prove that's impossible; you're trying to prove that it implies $Ha=Hb$. – Greg Martin May 17 '14 at 04:50
  • Go from $a=h_1^{-1}h_2b,$ to $,Ha=Hb$. – anon May 17 '14 at 05:10

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If you want to proof that by contradiction, you have to assume that $Ha\cap Hb\notin\lbrace \emptyset, G \rbrace$.

Going on from where you got stuck: $a=h_1^{-1}h_2b$ implies $Ha\subseteq Hb$ and multiplying from the left with $h_2^{-1}h_1$ gives $b=h_2^{-1}h_1a$ and therefore $Hb\subseteq Ha$. Combining the two inclusions gives $Ha=Hb$. Contradiction.