# An example of a ring R and a subring R' with R' not an ideal of R

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?

• See also the discussion here. – Dietrich Burde May 5 '15 at 18:24
• The second condition is for a subring. For a (two-sided) ideal, the condition is: For all $a\in I$, and all $\, b\in\color{red}R, …$. – Bernard May 5 '15 at 18:24
• The integers are a subring of the rationals, or reals. – André Nicolas May 5 '15 at 18:30
• 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? – user121615 May 5 '15 at 18:32
• 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? – user121615 May 5 '15 at 18:36