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I have to say that this article http://arxiv.org/abs/1506.00887 has at least one mistake.

Lemma 3.5, Counterexample - take group G=sequences$G=$ sequences of integer numbers $=\mathbb{Z}^\mathbb{N}$.

l$\ell$ and m$m$ are right and left shift respectively.

m*l=idThen $m\circ\ell=id$, so Image(e)=G$Image(e)=G$.

I have to say that this article http://arxiv.org/abs/1506.00887 has at least one mistake

Lemma 3.5 Counterexample - take group G=sequences of integer numbers

l and m are right and left shift respectively

m*l=id, so Image(e)=G

I have to say that this article http://arxiv.org/abs/1506.00887 has at least one mistake.

Lemma 3.5, Counterexample - take group $G=$ sequences of integer numbers $=\mathbb{Z}^\mathbb{N}$.

$\ell$ and $m$ are right and left shift respectively.

Then $m\circ\ell=id$, so $Image(e)=G$.

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Tomas Paul
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I have to say that this article http://arxiv.org/abs/1506.00887 has at least one mistake

Lemma 3.5 Counterexample - take group G=sequences of integer numbers

l and m are right and left shift respectively

m*l=id, so ImImage(e)=G

I have to say that this article http://arxiv.org/abs/1506.00887 has at least one mistake

Lemma 3.5 Counterexample - take group G=sequences of integer numbers

l and m are right and left shift respectively

m*l=id, so Im(e)=G

I have to say that this article http://arxiv.org/abs/1506.00887 has at least one mistake

Lemma 3.5 Counterexample - take group G=sequences of integer numbers

l and m are right and left shift respectively

m*l=id, so Image(e)=G

Source Link
Tomas Paul
  • 181
  • 1
  • 4

I have to say that this article http://arxiv.org/abs/1506.00887 has at least one mistake

Lemma 3.5 Counterexample - take group G=sequences of integer numbers

l and m are right and left shift respectively

m*l=id, so Im(e)=G