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For G a group and H a subgroup of G, I'm required to show that $|aH| = |bH|$ for all $a,b\epsilon G$

Proof: Define a function $f:aH\to bH$ such that $f(ah) = bh$ for some $h\epsilon H$ Then we show, f is 1-1 and onto, i.e. f is a bijection 1) f is 1-1: let $f(ah)=f(h')$ then $bh = bh'$ multiplying by $b^{-1}$ to the left we get $ h= h'$ Hence, f is injective

2) I have to show that f is onto but I don't know how. Can anyone let me know as well as verify the aforementioned proof for injectivity of f?

Alea
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1 Answers1

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One-one: If $f(ah)=f(ah')$, then $bh=bh'$ and the cancellation property gives $h=h'$.

Onto: For all $bh\in bH$, $f(ah)=bh$. This covers every element of $bH$.

Riley
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  • Is this statement sufficient? Also, is my injectivity proof alright and would this be good enough to conclude that $|aH| = |bH|$ – Alea Dec 07 '17 at 21:01
  • I think your proof should've stated $f(ah)=f(ah')$ instead of $f(ah)=f(h')$. Then the rest works out. Since $f$ is bijective, the cosets must have the same cardinality. This is a basic property of set theory. – Riley Dec 07 '17 at 21:03
  • Thank you! Yes, that was a typing error. Thanks! – Alea Dec 07 '17 at 21:06