# Some isomorphism conditions

Here I want to ask some true or false problems regarding isomorphisms (and if false, is there some extra condition to make it true). I do not what is the correct title for these problem. And I also want some brief proofs if possible.

1. Let $R$ and $R'$ be two rings with $|R| =|R'|< \infty$. And each proper ideal in $R$ is isomorphic to some ideal in $R'$. Then is it true that $R\cong R'$? (and if false, is there some extra condition to make it true).

2. Let $G$ and $G'$ be two groups with the same order ($< \infty$). If their abelianizations are isomorphic, is it true that $G\cong G'$.

3. Let $\mathcal{D}$ be a subcategory of a category $\mathcal{C}$. (1) If $u$ is an isomorphism of $\mathcal{D}$, is $u$ an isomorphism of $\mathcal{C}$? (2) If $v$ is an isomorphism of $\mathcal{C}$, is $v$ an isomorphism of $\mathcal{D}$? (and if false, is there some extra condition to make it true).

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What does it mean for two ideals to be isomorphic? – Qiaochu Yuan Apr 29 '11 at 4:53
3. (1) is obviously true; (2) make $\mathcal{D}$ a full subcategory. 2. there are simple groups of the same order. It remains to answer 1, which I leave to the algebraists after you clarified Qiaochu's question. – t.b. Apr 29 '11 at 4:57
@Qiaochu Yuan: I mean there exists ring homomorphisms $\psi: I\to I'$ and $\phi: I'\to I$ such that $\phi\circ\psi = id$ and $\psi\circ\phi = id$. – Junyu Apr 29 '11 at 4:59
(1) is not true. take $R=\mathbb{Z}/4\mathbb{Z}$ and $R'=\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. The only proper ideals of R are the zero ideal and $\mathbb{Z}/2\mathbb{Z}$ and they both appear in R'. Also, you might want to ask about proper subrings and not ideals (though it is still not true, but might be with extra conditions) – Prometheus Apr 29 '11 at 5:00
@user5980: what do you mean by a ring homomorphism between two ideals? (Is this a homomorphism of not necessarily unital rings, i.e. a rng homomorphism?) – Qiaochu Yuan Apr 29 '11 at 5:39

The answer to #2 is no. For example, the dihedral group $D_4$ and the quaternion group $Q_8$ both have order $8$ and abelianization $(\mathbb{Z}/2\mathbb{Z})^2$. (This implies, among other things, that they have the same character table.) Group theory would be very boring if anything like this was true.