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Let $N,H$ two subgroup of a group $G$ such that at least one is normal. By Surb answer here, $NH$ is the smallest group that contain $N$ and $H$. But if $I$ and $J$ are ideal, they are also group for $+$, therefore we should have $I\cap J\leq I,J\leq IJ$ (where $A\leq B$ means $A$ subgroup of $B$). So why with ideal is different whereas they are also groups ?

Because if $I$ and $J$ are ideal, then $IJ\subset I\cap J$, whereas by what I said before, we should have $I\cap J\leq IJ$.

P.S: I know how to show that $IJ\subset I\cap J$ so it's not my question ! I just want to understand the subtlety of this inclusion.

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Rick
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  • No, you say $I,J$ are groups for $+$, not for $\cdot$. Hence all we can from group theory is $I,J\subset I \color{red}{+} J$. – Bernard Oct 18 '15 at 10:24
  • You look to make confusion between the inclusion as set, and subgroup ! It has no sense to say that $I\cap J$ is a subgroup of $IJ$ since $IJ$ it's even not a group. But since $(I,+)$ and $(J,+)$ are group, you have that $I\cap J\leq I+J$. – Surb Oct 18 '15 at 10:24

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As the other commenters have noted, I think your confusion stems from the fact that $NH$ in the group sense isn't the same as $IJ$ in the ideal sense. The group corresponding to $NH$ for $I$ and $J$ should be $I+J$, since these things are groups in the additive (not multiplicative) sense. $IJ$ is a different sort of object, which we get from multiplication (which won't be a group operation for $I$ or $J$).

Your reasoning breaks when you claim that $J\subset IJ$. This doesn't follow from group theory, and in general isn't true. Take, for example, $I=J=2\mathbb{Z}$ in the integers. The problem is that $I$ might not contain a multiplicative identity 1. It'll only have the identity 0 in the group sense, which doesn't help us when we consider the multiplicative set $IJ$.

Josh Keneda
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