An analog of the converse of Lagrange's theorem for rings

I was wondering if there exists an analogue of the converse of Lagrange Theorem for abelian groups (for any divisor of the order of the finite abelian group, there is a subgroup of that order) for rings. More precisely, I know that if you have a finite rng $$R$$, then you $$R$$ is the direct product of the Sylow subgroups of $$(R,+)$$, which are subrngs. Now, is it true that every rng of prime power order contains subrng for every divisor of the order?

• Did you look at any nontrivial examples? What examples of rings (or rngs) of prime power order do you know? There are at least two easy classes of examples. Commented Aug 1 at 17:14
• @QiaochuYuan Are you thinking about polynomials and matrices over let's say a field of prime order $p$? Commented Aug 1 at 21:16
• No, even simpler than that. The desired statement is already false for the finite fields $\mathbb{F}_{p^n}$. Commented Aug 1 at 21:51
• @QiaochuYuan Ok, so the thing is that if I take a field of order $p^3$, then it does not contain subfields of order $p^2$, so even subrngs of the same order because every subrng is a subfield Commented Aug 2 at 6:24
• Yes, that's what I had in mind. Commented Aug 2 at 6:29