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Let $G$ be a finite abelian group. Then is there a $\cup$-like operation we will call group union such that it distributes over subgroup intersection?

Let $\mathcal{H}(G)$ be the set of subgroups of $G$, let $\cup$ be our group union and $\cap$ the usual group intersection.

Then does there exist a union operator $\cup$ such for all $H,K,L \leqslant \mathcal{H}(G)$, the following hold:

$$ H\cap(K \cup L) = (H\cap K) \cup (H\cap L), \\ H\cup(K \cap L) = (H\cup K) \cap (H \cup L) $$

If not, then are there any such that just one of the above holds?

My attempt: If you define $H\hat{\cup} K$ to be the smallest subgroup containing the union. Then it's easy to show that $H\hat{\cup}(K\cap G) \subset (H\hat{\cup}K)\cap(H\hat{\cup}G)$. But to show the opposite is tricky, i.e. I'm stuck.

If the opposite inclusion isn't true then there's $x \notin H$ (since if there were we'd be done.), and either, without loss of generality, (case 1) $x\in K$ and $\notin G$, or, (case 2) $x\notin$ any of the subgroups but is a generated element.

(Case 1): If there's such an $x$ then $x = k = hg$ for some $k,h,g$ in their respective groups. This is close enough to where I give up, so I'll stop. Any ideas?

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Try $H, K, L$ the three different subgroups of order $2$ of symmetric group $S_3$ on three elements. – j.p. Oct 2 '13 at 10:33
@j.p. You mean that's a counter example to the inclusion involving $\hat{\cup}$ ? – Enjoys Math Oct 2 '13 at 19:18
What do you get if you use this three subgroups of $S_3$ in your attempt (where $L$ is called $G$)? – j.p. Oct 3 '13 at 17:16

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