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Let $S$ be a set, and $*$ an associative binary operation on $S$. Suppose there is an element $e\in S$ such that

($1$) $e*x=x$ and $x*e=x$ for all $x\in S$.

(a) Prove that there is a unique element $e$ satisfying property $(1)$.

(b) Let $T=\{x\in S$ | there exist $y$, $z$ $\in S$ such that $y*x=e=x*z$}. Prove that $(T, *)$ is a group.

I understand ($1$) perfectly. However, I am completely lost on ($2$). I don't know how to proceed to verify the group axioms. Any help/solutions will be greatly appreciated.

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  • $\begingroup$ Please do not edit your question after it has been answered. $\endgroup$ – user133281 Aug 27 '14 at 23:14
  • $\begingroup$ Maybe no one is doing this because this site is not intended for tutoring (or solving problems on demand)... $\endgroup$ – colormegone Aug 28 '14 at 2:08
  • $\begingroup$ "Hurry! Why isn't anyone doing this?" Seriously!?!? People who ask questions like this are going to drive me away from this site eventually. There is no reason any question deserves to get an answer--if anyone responds at all, consider it like an incredible gift! Don't be demanding. $\endgroup$ – apnorton Aug 28 '14 at 2:15
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    $\begingroup$ @user133281 You should roll back the edits when this happens. $\endgroup$ – user1729 Aug 28 '14 at 8:13
  • $\begingroup$ (Also, I am voting to re-open this question because the original question was fine. It is the self-defacing that is the issue.) $\endgroup$ – user1729 Aug 28 '14 at 8:14
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It is easy to show that $e \in T$ (since $e \ast e = e$) and that $e$ acts as a unit element. Furthermore, $T$ contains inverses of its elements. By definition, the operation $\ast$ is associative. The most tricky part is showing that $T$ is closed under $\ast$. Suppose that $a$, $b \in T$. Then we have to show that $a \ast b$ is in $T$ as well, thus we have to show that there exists $c,d \in T$ such that $a \ast b \ast c = e$ and $d \ast a \ast b = e$. We know that there is an $x \in T$ such that $b \ast x = e$ and that there is a $y \in T$ such that $a \ast y = e$. Then we have $a \ast b \ast (x \ast y) = e$, so we can take $c = x \ast y$. Analogously, we can find a suitable choice for $d$. (Note that it is not required that $x \ast y$ is itself in $T$, only that it is in $S$.)

Edit:

Why is $e$ in $T$? Because there exist $y,z \in S$ such that $y \ast e = e = e \ast z$. One can take $y=z=e$ since $e \ast e = e$ (applying $e \times x = x$ with $x = e$).

Why contains $T$ inverses of its elements? Suppose $a \in T$. Then there are $b,c \in S$ with $a \ast b = e$ and $c \ast a = e$. This means that $c = c \ast e = c \ast (a \ast b) = c \ast a \ast b = (c \ast a) \ast b = e \ast b = b$, hence $b = c$. So we have $a \ast b = e$ and $b \ast a = e$. Note that this implies that $b \in T$, and therefore $b$ is an inverse of $a$.

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  • $\begingroup$ What is T? Where is it defined? $\endgroup$ – Geoff Robinson Aug 27 '14 at 23:10
  • $\begingroup$ This question was edited. The original question mentioned a set $S$ equipped with a binary associative operation $\ast$, an element $e \in S$ with $x \ast e = e \ast x = x$ for all $x \in S$, and asked to show that the set $T = \{x \in S: y \ast x = e, x \ast z = e \mbox{ for some } y,z \in S\}$ is a group under $\ast$. $\endgroup$ – user133281 Aug 27 '14 at 23:13

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