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Let $X,Y,Z$ be subspaces of $V$ so that $X$ is a subspace of $Y$. Prove that $Y\cap (X+Z)=X+(Y\cap Z)$

I know that I need to prove that $Y\cap (X+Z)\subseteq X+(Y\cap Z)$ and $X+(Y\cap Z)\subseteq Y\cap (X+Z)$ but I´m having a hard time.

Can you lend me a hand please? I would really appreciate it

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  • $\begingroup$ so what precisely have you tried? $\endgroup$ Sep 4, 2014 at 12:49
  • $\begingroup$ I give a $v\in Y\cap (X+Z)$ so $v\in Y$ and $v\in (X+Z)$ and the latter one implies by definition that $v=v_1+v_2$ where $v_1\in X$ and $v_2\in Z$ but I don´t know what to do from here $\endgroup$
    – user128422
    Sep 4, 2014 at 12:54
  • $\begingroup$ okay I have added an answer to help you out. I just realized that you already mentioned the general hint, but anyways... $\endgroup$ Sep 4, 2014 at 13:02

4 Answers 4

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This is called a modular law. It holds for subgroups of any abelian (additive) group, thus it holds also for submodules of a module, and in particular for subspaces of a vector space. I prefer to write it in the form that is easy to remember: $\newcommand{\inters}{\cap}$

Let $A$, $X$, $B$ be subgroups of an abelian group. If $A\subseteq B$, then $A+(X\inters B)=(A+X)\inters B$.

You see -- you just move the pair of parentheses.

Proof. $~$Since $A\subseteq A+X$ and $A\subseteq B$, we have $A\subseteq (A+X)\inters B\,$; since $X\subseteq A+X$, we have also $X\inters B\subseteq(A+X)\inters B\,$; it follows that $A+(X\inters B)\subseteq (A+X)\inters B\,$. In order to prove the reverse inclusion, we consider any $y\in(A+X)\inters B\,$, that is, $y=a+x=b$ for some $a\in A$, some $x\in X$, and some $b\in B$. Then $x=b-a\in B$ because $b\in B$ and $a\in A\subseteq B$, thus $x\in X\inters B$, whence $y=a+x\in A+(X\inters B)\,$.$~$ Done.

The proof is essentially the same as the one given by Deuteu, just cleaner and clearer. $\newcommand{\coll}{\mathcal}$ $\newcommand{\Inters}{\bigcap}$

Let $\coll{L}$ be the set of all subgroups of an abelian group (or of all submodules of a module, or of all subspaces of a vector space). When $\coll{L}$ is ordered by inclusion, it becomes a lattice: any $X,Y\!\in\coll{L}$ have the greatest lower bound $X\inters Y$ and the least upper bound $X+Y\!$. (Actually every indexed family of members $X_i$ of $\coll{L}$ has the greatest lower bound $\Inters_iX_i$ and the least upper bound $\sum_iX_i$
-- the lattice $\coll{L}$ is complete). The lattice $\coll{L}$ is a modular lattice because (you guessed it)
it obeys the modular law. See the article "Modular lattice" in Wikipedia (for starters).

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As noted, to prove an equality of sets one must prove each side is a subset of the other one.

First, we prove the $\subseteq$ side, i.e. $Y \cap (Z+X) \subseteq X + (Y \cap Z)$:

Take any $a \in Y \cap (Z+X)$. That means $a$ is in both $Y$ and $(Z+X)$. The question is: is $a \in X$?

  • if $a \in X$, that means that it certainly is in $X+(Y\cap Z)$.
  • if $a\notin X$, then it must be in $Z$ (since it is in $Z+X$). Thus, $a$ is in both $Y$ and $Z$, which means it is in $Y\cap Z$ and therefore it is also in $X+(Y\cap Z)$.

Second, prove that $X+(Y\cap Z) \subseteq Y\cap (Z+X)$:

Take any $a \in X+(Y\cap Z)$. Then, $a$ is either in $X$ or in the intersection $Y\cap Z$. Since $X$ is a subset of $Y$, we have that $a$ must be in $Y$. Now, the same question: is $a\in X$?

  • if $a\in X$, it is also in $Z+X$ and we are done.
  • if $a \notin X$, it must be in $Y\cap Z$, which means it certainly is in $Y\cap (Z+X)$. $\Box$
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A collection of hints:

Hint: $X + Y = Y$

General hint: if you try to show $A \subset B$ then take a generic element $a \in A$ and show $a \in B$.

Hint about elements: a generic element of the LHS is of the form: $Y \ni y=x+z \in X+Z$ (it lies in the intersection). When does this happen to be in the RHS? In which subspace(s) lives $z = y-x$? (hint: it is not only $Z$)

Another hint about elements: if $x + r$ is a generic element on the RHS, then in particular it is an element in $X+Z$. So it suffices to show that it is also in $Y$. Why is that?

If you use those hints you should be able to solve the exercise yourself - hope it helps.

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As you said we can do it by double inclusion there is :

If $w \in X+(Y\cap Z)$, we have $w=x+\mu$ with $x\in X$ and $\mu \in (Y\cap Z)$
- $X \subseteq Y$ so $x \in Y$ thus $x,\mu \in Y$ so $w \in Y$
- $\mu \in Z$ so $w \in (X+Z)$
In conclusion $w \in Y\cap (X+Z)$

If $v \in Y\cap (X+Z)$, $v=\alpha \in Y$ and $v=\beta + \gamma$ with $\beta \in X$ and $\gamma \in Z$
$\gamma = \alpha - \beta$ and $\beta \in X \subseteq Y$ so $\gamma \in Y$ thus $\gamma \in Y \cap Z$
In conclusion $v=\beta_{\in X} + \gamma_{\in Y\cap Z}$ so $v\in X+(Y\cap Z)$

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