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

• Does $\leq$ symbol here, means an Ideal or Normal subgroup? Also, Please clearly state whether you want the answer from the perspective of group theory or ring theory. Also, It's not clear that you are using the Third isomorphism for rings or groups or modules or field or General. – Kumar Jul 16 at 6:16
• Not in general. Suppose $Y$ is a group and $X_2 \lhd Y$ is a normal subgroup that contains some subgroup $X_1$ that is not normal in $X_2$. Then $X_1 \cap X_2 = X_1$ is of course normal in itself but is not normal in $X_2$, so certainly not normal in $Y$. For a concrete example, take $Y = S_4$, $X_2 = A_4$, and $X_1 = \{1, (12)(34)\}$. – Bungo Jul 16 at 6:17
• @Kumar, I edited my question. – se-hyuck yang Jul 16 at 7:25

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 $$X_2$$.