Normal subgroup and ideal inclusion I'm a student who just started  abstract algebra.
There is an easy question for you.
Let $X_1 \le Y$ which means $X_1$ is a subobject(subring or subgroup) of   $Y$
Also let $X_2$ bes a (normal or ideal) of   $Y$
According to the (group or ring) $2nd$ isomorphism theorem, We can easily pull out the conclusion that $X_1 \cap X_2$ is an ideal of or normal in   $X_1$. 
But Question is 
If the sets $X_1$ and $X_2$ are not contained in each other
(I.E. There aren't case that $X_1  \subset  X_2$ or $X_2  \subset  X_1$)
And not disjoint from each other then...
$(1)$ 
Considering group case, is $X_1 \cap X_2$ a normal subgroup of  $X_2$?
Considering ring case, is $X_1 \cap X_2$   an ideal of   $X_2$?
$(2)$
Considering group case, ie $X_1 \cap X_2$ is a normal subgroup of  $Y$?
Considering ring case, Does $X_1 \cap X_2$ is a ideal of the $Y$?
Whenever I try proving n (1) and (2), they look like true.
But I don't have confidence that both of them bre true.
Any help would be appreciated.
Thank you.
Thanks to the @bungo, I add more conditions
 A: The answer to each question is "no".
Consider $G$ to be any group/ring and $H$ to be a non-normal subgroup/non-ideal subring. I could say take $X_1=H$ and $X_2=G$ and be done but you additionally want to have that they are not contained in each other. We do that by extending $G$: you take another nontrivial group/ring $G'$ and consider $G'\times G$. Now you take $X_1=G'\times H$ and $X_2=\{0\}\times G$ (I denote the neutral element by "$0$" here to be consistent between groups and rings, but in the case of groups I don't assume that $G'$ is abelian, it is arbitrary). Note that $X_2$ is normal/ideal in $G'\times G$. The intersection $X_1\cap X_2$ is still "the same" as in the non-extended case but they are not contained in each other this time.
Now since $X_1\cap X_2$ is not normal/not ideal in $X_2$ then it cannot be in any superobject containing $X_2$.
Side note: "And not disjoint from each other" algebraic subobjects are never disjoint (each of them contains the neutral element). Did you mean nontrivial intersection? You don't need to assume that because the trivial subobject is always normal/ideal in anything.
Side note 2: in the case of unital rings there are subtleties because typically a subring is assumed to contain the multiplicative identity. And thus a proper ideal is never a subring. So we need a more general definition of a subring for that to even make sense. And I'm not sure if everything works correctly with this approach (it probably does).
