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And also, another thing, I'm curious about. They say that an Ideal is the analogue of the Normal subgroup in group theory, but that confuses me.

Let a Group be G. Let a subgroup be H. H is normal in G if and only if, $gH = Hg$, for all $g$ in $G$.

An Ideal:

Let $<R,+,*>$ Be a ring with set $R$ and the usual operations of addition and subtraction. Let $I$ be a subset of $R$. We say that $I$ is an ideal if it forms a subgroup $<I,+>$ of $<R,+>$ under addition. And also, for $a,b$ that is in $I$:

$a*b$ is in $I$ and $b*a$ is in $I$.

So a subgroup $H$ is normal if and only if for all elements $aH = Ha$, but a SET is an ideal if the following conditions hold. I don't understand the analogue of the two. An ideal requires that multiplication be defined and in our set I. Normal subgroups require something different. What am I missing?

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  • $\begingroup$ See also the discussion here. $\endgroup$ – Dietrich Burde May 5 '15 at 18:24
  • $\begingroup$ The second condition is for a subring. For a (two-sided) ideal, the condition is: $$$$For all $a\in I$, and all $\, b\in\color{red}R, …$. $\endgroup$ – Bernard May 5 '15 at 18:24
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    $\begingroup$ The integers are a subring of the rationals, or reals. $\endgroup$ – André Nicolas May 5 '15 at 18:30
  • $\begingroup$ I know but an Ideal is a set, not a ring. While a Normal subgroup is a group. One has an operation defined on it, the other doesn't. How can that be analogous? Is the subring that it forms the analogy? $\endgroup$ – user121615 May 5 '15 at 18:32
  • $\begingroup$ What I don't understand is why they just don't say that I is a subring? Why define it as a set? They must have a reason.Would a good way to look at an ideal be that an Ideal is a set that can form a subring? $\endgroup$ – user121615 May 5 '15 at 18:36

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