Let L be any non-empty language over an alphabet Σ. Show that $L^2$ ⊆ $L^3$ if and only if λ ∈ L.

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I have this question in my Theoretical Computer Science class on the topic of Automata and Formal Languages.

$$ λ ∈ L \iff L^2 ⊆ L^3$$

I thought of showing it by proving the contrapositive but I was only able to complete half of the proof (since it is a bidirectional proof) and on top of that I am not even sure it is correctly written in a formal form.

I wanted to prove that if

$$λ ∉ L \implies L^2 \nsubseteq L^3$$

But since it is a bidirectional proof, I must also show that

$$L^2 \nsubseteq L^3 \implies λ ∉ L$$

right?

This is what I have so far:

1) $λ ∉ L \implies L^2 \nsubseteq L^3$

$ \lambda \notin L \implies \forall w \in L, n_0(w) \geq 1$
$ L^2 = L * L \implies \forall w_2 \in L^2, n_0(w_2) \geq 2 $
$ L^3 = L * L * L \implies \forall w_3 \in L^3, n_0(w_3) \geq 3 $
$\forall w_3 \in L^3 \land \forall w_2 \in L^2, n_0(w_3) > n_0(w_2)$
$ \therefore L^2 \nsubseteq L^3$

So first of all, is this correct?

Secondly, how would I go about showing the second part of my proof where
$$L^2 \nsubseteq L^3 \implies λ ∉ L$$

Thank you!

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As Orest mentioned in the comments, $$\forall w_3 \in L^3 \land \forall w_2 \in L^2, n_0(w_3) > n_0(w_2)$$ is false. To give a correct proof of this direction, instead take $w_1 \in L$ to be a word of minimal length and show the concatenation $w_1w_1$ is in $L^2$ but not in $L^3$

For the other direction, assume $\lambda \in L$ and concatenate any word in $L^2$ with $\lambda$ to find a word in $L^3.$