## Finite groups G in which for every subgroup H of G, Hom(G,H) contains an element that is identity on H

Posted: July 26, 2020 in Elementary Algebra & Number Theory; Problems & Solutions, Groups and Fields

Problem. Let $G$ be a finite group with $|G| > 1$ and suppose that $G$ satisfies the following condition:

$(*)$    for every subgroup $H$ of $G$ there exists a group homomorphism $f: G \to H$ such that $f(h)=h, \ \ \ \forall h \in H.$

Prove that $G$ is isomorphic to a direct product of cyclic groups of prime orders.

Solution. We first prove a claim.

Claim. Every subgroup of $G$satisfies $(*).$

Proof. Let $K$ be a subgroup of $G$ and let $H$ be any subgroup of $K.$ We’re given that there exists a group homomorphism $f: G \to H$ such that $f|_H$ is the identity map. Then clearly $g:=f|_K: K \to H$ is a group homomorphism and $g|_H$ is the identity map. So $K$ satisfies $(*)$ too.

Now, back to the solution. One way to solve the problem is by induction over $|G|.$ There’s nothing to prove if $|G|=2.$ Now consider two cases.

Case 1: $G$ is simple, i.e. $G$ has no non-trivial normal subgroup. Suppose that $G$ has a non-trivial subgroup $H.$ Since $G$ satisfies $(*),$ there exists a group homomorphism $f: G \to H$ such that $f|_H$ is the identity map. Then $K:=\ker f$ is a normal subgroup of $G$ and so either $K=(1)$ or $K=G.$ If $K=(1),$ then $G=H$ and if $K=G,$ then $H=(1),$ which are both false because $H$ is non-trivial. So $G$ has no non-trivial subgroup and hence it’s a cyclic group of prime order. That solves the problem in this case.

Case 2: $G$ is not simple. So $G$ has a non-trivial normal subgroup $H.$ Since $G$ satisfies $(*),$ there exists a group homomorphism $f: G \to H$ such that $f|_H$ is the identity map. Let $K:=\ker f.$ Note that since $H$ is non-trivial, $K$ is non-trivial too. Now, since $f$ is onto, $G/K \cong H$ and so $|G|=|K| \cdot |H|.$ Let $x \in K \cap H.$ Then $1=f(x)=x$ and so $K \cap H=(1).$ Thus $|KH|=|K| \cdot |H|=|G|$ and hence $G=KH.$ Therefore, since $K,H$ are normal and $K \cap H=(1),$ we have $G \cong K \times H.$ The solution is now complete because, by the above claim, both $H,K$ satisfy $(*)$ and hence, by our induction hypothesis, both are a direct product of cyclic groups of prime orders and hence $G \cong K \times H$ is also a direct product of cyclic groups of prime orders. $\Box$

## The set of anti-derivatives of a continuous function as an abelian group

Posted: July 17, 2020 in Elementary Algebra & Number Theory; Problems & Solutions, Groups and Fields

Problem. Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function and let $G$ be the set of all anti-derivatives of $f.$ Let $c \in \mathbb{R}$ and define the binary operation $*: G \times G \to G$ by $u *v=u+v(c),$ for all $u,v \in G.$ Show that $(G,*)$ is an abelian group isomorphic to $(\mathbb{R},+).$

Solution. If $u,v \in G,$ then $u-v$ is a constant function (because both $u,v$ are anti-derivatives of $f$) and so

$u-v=u(c)-v(c). \ \ \ \ \ \ \ \ \ \ \ \ (\dagger)$

1) $*$ is well-defined because if $u \in G,$ then $(u+v(c))'=u'=f$ and so $u*v=u+v(c) \in G.$

2) $*$ is commutative because, by $(\dagger),$

$u*v=u+v(c)=v+u(c)=v*u.$

3) $*$ is associative because It’s easy to see that $(u*v)*w=u*(v*w)=u+v(c)+w(c)$ for all $u,v,w \in G.$

Now fix an element $u_0 \in G.$

4) $u_0-u_0(c)=1_G$ because if $u \in G,$ then

$u*(u_0-u_0(c))=u+(u_0-u_0(c))(c)=u+u_0(c)-u_0(c)=u$

and so, by 2), we also have $(u_0-u_0(c))*u=u,$ proving that $u_0-u_0(c)$ is the identity element of $G.$ Note that $1_G(c)=0.$

5) $u^{-1}=1_G-u(c)$ because $(1_G-u(c))*u=1_G-u(c)+u(c)=1_G$ and we’re done by 2).

So we have shown that $(G,*)$ is an abelian group.

To show that $(G,*) \cong (\mathbb{R},+),$ we define the map

$\phi:(G,*) \to (\mathbb{R},+), \ \ \ \ \ \ \ \phi(u)=u(c).$

6) $\phi$ is a group homomorphism because

$\phi(u*v)=\phi(u+v(c))=u(c)+v(c)=\phi(u)+\phi(v).$

7) $\phi$ is injective because $\phi(u)=\phi(v)$ gives $u(c)=v(c)$ and so, by $(\dagger),$ we have $u=v.$

And finally

8) $\phi$ is surjective because if $r \in \mathbb{R}$ and $u:=1_G+r \in G,$ then $\phi(u)=u(c)=1_G(c)+r=r. \ \Box$

## Finite fields F in which the subfield generated by {x^3: x \in F} = F

Posted: July 8, 2020 in Elementary Algebra & Number Theory; Problems & Solutions, Groups and Fields
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Problem. Let $F$ be a finite field and let $K$ be the subfield of $F$ generated by the set $\{x^3: \ \ x \in F\}.$ Show that either $K=F$ or $|F|=4.$

Solution. Suppose that $K \ne F.$ We have $|F|=p^n$ for some prime number $p$ and some positive integer $n.$ So $|K|=p^d$ for some $d \mid n.$ Let $F^{\times}, K^{\times}$ be the multiplicative groups of $F,K,$ respectively. Since $x^3 \in K^{\times}$ for all $x \in F^{\times}$ and $K^{\times} \ne F^{\times},$ we have $|F^{\times}/K^{\times}|=3.$ So

$\displaystyle 3=|F^{\times}/K^{\times}|=\frac{|F^{\times}|}{|K^{\times}|}=\frac{p^n-1}{p^d-1}=p^{n-d}+p^{n-2d} + \cdots + p^d+1$

and thus $p=2, n-d=d=1$ because $p \ge 2.$ So $|F|=4$ and the solution is complete. $\Box$

Exercise. Let $F$ be the field of order $4,$ and let $K$ be the subfield of $F$ generated by $\{x^3: \ x \in F\}.$ Show that $|K|=2.$

## Matrices A,B satisfying AB-BA=c(A-B) for some non-zero number c, have the same eigenvalues

Posted: July 2, 2020 in Elementary Algebra & Number Theory; Problems & Solutions, Linear Algebra
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Problem. Let $A,B$ be two non-zero elements of $M_n(\mathbb{C})$ such that $AB-BA=c(A-B)$ for some $0 \ne c \in \mathbb{C}.$ Show that $A,B$ have the same eigenvalues.

Solution. Let $I \in M_n(\mathbb{C})$ be the identity matrix. Let $p(x),q(x) \in \mathbb{C}[x]$ be the characteristic polynomials of $A,B,$ respectively. See that $AB-BA=c(A-B)$ gives

$(xI-A)((x+c)I-B)=(xI-B)((x+c)I-A)$

and thus, taking the determinant, we get

$p(x)q(x+c)=q(x)p(x+c).$

Therefore $r(x+c)=r(x),$ where $\displaystyle r(x)=\frac{p(x)}{q(x)},$ i.e. $c \ne 0$ is a period of the rational function $r(x),$ which is non-sense unless $r(x)$ is constant. So $r(x)$ is a constant function and hence, since $p(x),q(x)$ are both monic, $p(x)=q(x)$ and the result follows. $\Box$

Exercise. Let $f$ be a non-constant rational function over $\mathbb{C}.$ Show that $f$ cannot be periodic. Does that remain true if we replace $\mathbb{C}$ with an arbitrary field?

## Rings in which the equation ax + xa = 1 has a solution for every non-zero element a

Posted: June 6, 2020 in Noncommutative Ring Theory Notes, Simple Rings

In this post, we take a look at the rings $R$ with $1$ that satisfy the following property

$\forall a \in R \setminus \{0\}, \ \exists x \in R: \ \ \ ax+xa=1. \ \ \ \ \ \ \ \ \ \ \ \ (*)$

In other words, rings in which the equation $ax+xa=1$ has a solution for every $0 \ne a \in R.$

Remark 1. It is clear that a commutative ring satisfies $(*)$ if and only if it is a field of characteristic $\ne 2$ because in this case $(*)$ becomes $2ax=1.$ It is also clear that any division ring $R$ of characteristic $\ne 2$ satisfies $(*)$ because if $0 \ne a \in R$ and $x=2^{-1}a^{-1},$ then $ax+xa=1.$

Remark 2. The equation $ax+xa=1$ could have more than one solution in a ring. For example, in the division ring of real quaterninos $\mathbb{H}=\{\alpha+\beta i + \gamma j + \delta k, \ \ \alpha, \beta, \gamma, \delta \in \mathbb{R}\},$ choose $a=i+j$ and $x=\beta i-\left(\frac{1}{2}+\beta\right)j,$ where $\beta$ is any real number, and see that $ax+xa=1.$
However, if $a=\alpha+\beta i + \gamma j + \delta k \in \mathbb{H}$ and $\alpha \ne 0,$ then the equation $ax+xa=1$ has the unique solution $x=2^{-1}a^{-1}$ (why ?).

We now show that any ring that satisfies $(*)$ is a simple domain of characteristic $\ne 2.$

Proposition. Let $R$ be a ring and suppose that $R$ satisfies $(*).$ Then $R$ is a simple domain and its center is a field of chracateristic $\ne 2.$

Proof. Let $I \ne (0)$ be an ideal of $R$ and choose $0 \ne a \in I.$ There exists $x \in R$ such that $ax+xa=1$ and so $1=ax+xa \in I$ implying that $I=R.$ So $R$ is a simple ring and thus its center is a field (see Remark 1 in this post!). For $a=1,$ there exists $x \in R$ such that $2x=ax+xa=1$ and so $2 \ne 0$ in $R,$ i.e. the characteristic of the center of $R$ is $\ne 2.$

It remains to show that $R$ is a domain. We’ll do that by first proving two claims.

Claim 1. If $a,x \in R$ and $ax+xa =1,$ then $a^2x=xa^2.$

Proof. $a^2x=a(ax)=a(1-xa)=a-axa=(1-ax)a=(xa)a=xa^2. \ \Box$

Claim 2. If $a \in R$ and $a^2=0,$ then $a=0.$

Proof.  Suppose, to the contrary, that there exists $0 \ne a \in R$ such that $a^2=0$ and let $x \in R$ be such that $ax+xa=1.$ Then

$a=a(ax+xa)=axa, \ \ \ (ax)^2=ax \ \ \ \ \ \ \ \ \ \ \ \ (1)$

and, as a result, $ax \ne 0.$ So there exists $y \in R$ such that

$axy+yax=1. \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

But then, by $(1)$ and Claim 1, $axy=(ax)^2y=y(ax)^2=yax$ and so by $(2), \ \ 2axy=1$ and hence $a=2a^2xy=0,$ contradiction! $\Box$

We are now ready to show that $R$ is a domain. Suppose that $ab=0$ for some $a,b \in R$ and $a \ne 0.$ To show that $R$ is a domain, we just need to prove that $b=0.$ We have $(ba)^2=b(ab)a=0$ and so, by Claim 2,

$ba=0. \ \ \ \ \ \ \ \ \ \ \ \ (3)$

Also, since $a \ne 0,$ there exists $x \in R$ such that $ax+xa=1$ and so

$b^2=b(ax+xa)b=(ba)xb+bx(ab)=0,$

by $(3).$ Hence, by Claim 2, $b=0$ and that completes the proof. $\Box$

Remark 3. Since an artinian domain is a division ring, it follows from the above problem and Remark 1 that an artinian ring satisfies $(*)$ if and only if it is a division ring of characteristic $\ne 2.$

Question. Is it true that a ring $R$ satisfies $(*)$ if and only if $R$ is a division ring of characteristic $\ne 2$ ? In other words, is there an example of a  ring that satisfies $(*)$ but the ring is not a division ring?

## G/Z(G) abelian & m odd imply that the set {g: g^m = 1} is a subgroup

Posted: March 24, 2020 in Elementary Algebra & Number Theory; Problems & Solutions, Groups and Fields
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Problem 1. Let $G$ be a group with the center $Z(G).$ Given an odd integer $m \ge 1,$ let

$H:=\{g \in G: \ \ g^m=1\}.$

Show that if $\displaystyle \frac{G}{Z(G)}$ is abelian, then $H$ is a normal subgroup of $G.$

Solution. It is clear that $1 \in H$ and $g \in H$ implies that $g^{-1} \in H.$ Also, if $x \in G$ and $g \in H,$ then $(xgx^{-1})^m=xg^mx^{-1}=xx^{-1}=1$ and so $xgx^{-1} \in H.$ Thus the only thing we need to prove is that $H$ is multiplicatively closed. So let $g_1,g_2 \in H.$ We want to show that $g_1g_2 \in H.$ Let

$g:=g_1g_2g_1^{-1}g_2^{-1}.$

Claim 1: $g \in Z(G).$

Proof. Since $\displaystyle \frac{G}{Z(G)}$ is abelian, $g_1^{-1}Z(G)$ and $g_2^{-1}Z(G)$ commute, i.e. $g_1^{-1}g_2^{-1}Z(G)=g_2^{-1}g_1^{-1}Z(G),$ and thus $gZ(G)= g_1g_2g_1^{-1}g_2^{-1}Z(G)=Z(G).$

Claim 2: $g_1^kg_2=g_2g_1^kg^k$ for all integers $k \ge 1.$

Proof. By induction over $k.$ By Claim 1, $g$ and $g_2g_1$ commute and so

$g_2g_1g=gg_2g_1=g_1g_2g_1^{-1}g_2^{-1}g_2g_1=g_1g_2$

and that proves the claim for $k=1.$ Now suppose the claim is true for $k.$ Then

$g_1^{k+1}g_2=g_1g_1^kg_2=g_1g_2g_1^kg^k=g_2g_1gg_1^kg^k=g_2g_1g_1^kgg^k=g_2g_1^{k+1}g^{k+1}$

and that completes the induction.

Claim 3: $g \in H.$

Proof. By Claim 2, $g^k=g_1^kg_2g_1^{-k}g_2^{-1}$ for all integers $k \ge 1$ and so, in particular, for $k=m$ we get

$g^m=g_1^mg_2g_1^{-m}g_2^{-1}=g_2g_2^{-1}=1.$

Claim 4: $(g_1g_2)^k=g_1^kg_2^kg^{-k(k-1)/2}$ for all integers $k \ge 1.$

Proof. The proof is by induction over $k.$ There’s nothing to prove for $k=1.$ Suppose now that the claim is true for $k.$ By Claim 2, $g_2g_1^k=g_1^kg_2g^{-k}$ and thus

$(g_1g_2)^{k+1}=g_1g_2(g_1g_2)^k=g_1g_2g_1^kg_2^kg^{-k(k-1)/2}=g_1g_1^kg_2g^{-k}g_2^kg^{-k(k-1)/2}$

$=g_1^{k+1}g_2^{k+1}g^{-k-k(k-1)/2}=g_1^{k+1}g_2^{k+1}g^{-k(k+1)/2},$

which completes the induction and the proof of the claim.

Now, if in Claim 4, we put $k=m,$ then we get

$(g_1g_2)^m=g_1^mg_2^mg^{-m(m-1)/2}=g^{-m(m-1)/2}, \ \ \ \ \ \ \ \ \ \ (*)$

because $g_1,g_2 \in H.$ But since $m$ is odd, $m$ divides $m(m-1)/2$ and hence $g^{-m(m-1)/2}=1$ because, by Claim 3, $g \in H.$ Therefore, by $(*), \ (g_1g_2)^m=1,$ i.e. $g_1g_2 \in H,$ and that completes the solution. $\Box$

Problem 2. Let $G$ be a non-abelian group of order $p^3,$ where $p$ is a prime number. Let $Z(G)$ be the center of $G$ and let $H:=\{g \in G: \ \ g^p=1\}.$ Show that

i) $|Z(G)|=p$

ii) if $p$ is odd, then $H$ is a normal subgroup of $G$ and $|H| \ge p$

iii) if $p=2,$ then $H$ is not always a subgroup.

Solution. i) Since $G$ is a non-abelian $p$-group, $1 < |Z(G)| < p^3=|G|$ and So either $|Z(G)|=p$ or $|Z(G)|=p^2.$ If $|Z(G)|=p^2,$ then $\displaystyle \frac{G}{Z(G)}$ would be a group of order $p,$ hence cyclic. But then $G$ would be abelian, which is non-sense. So $|Z(G)|=p.$

ii) By i), $\displaystyle \frac{G}{Z(G)}$ is a group of order $p^2$ hence abelian. So, by Problem 1, $H$ is a normal subgroup of $G.$ It is clear that $|H| \ge p$ because, by i), $Z(G) \subseteq H.$ You may also argue that $|H| > 1$ because, by Cauchy, $G$ has an element of order $p.$ Thus since $|H|$ divides $|G|=p^3,$ we must have $|H| \ge p.$

iii) Consider $D_8,$ the dihedral group of order $8$

$D_8=\langle a,b: \ \ \ a^2=b^4=1, \ \ aba=b^{-1}\rangle.$

Then $a^2=(ab)^2=1$ but $(aab)^2=(a^2b)^2=b^2 \ne 1$, i.e. in this case, $H$ is not closed under multiplication hence not a subgroup. $\Box$

Exercise. Let $R$ be a commutative ring with unity and let $G$ be the Heisenberg group over $R.$ Show that for any odd integer $m \ge 1,$ the set $\{g \in G: \ \ g^m=1\}$ is a normal subgroup of $G.$

## A very little remark on conjugate subgroups

Posted: February 28, 2020 in Elementary Algebra & Number Theory; Problems & Solutions, Groups and Fields
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Let $G$ be a group and let $H$ be a subgroup of $H.$ Let $g \in G.$ If $H$ is normal, then $gHg^{-1}=H.$ If $H$ is not normal but $gHg^{-1} \subseteq H$ and $H$ is finite, then again $gHg^{-1}=H$ because $|gHg^{-1}|=|H|.$ But what if $H$ is not normal, $gHg^{-1} \subseteq H$ and $H$ is not finite? Would that again imply $gHg^{-1}=H ?$ Of course not, here is an example.

Example. Consider the group $G:=\text{GL}(2,\mathbb{Q}),$ the group of $2 \times 2$ invertible matrices with rational entries. For any given $k \ge 1,$ consider the set

$\displaystyle H_k:=\begin{pmatrix}1 & k\mathbb{Z} \\ 0 & 1\end{pmatrix}.$

So $H_k$ is the set of all $2 \times 2$ matrices in the form $\displaystyle \begin{pmatrix}1 & kn \\ 0 & 1\end{pmatrix}$ for some integer $n.$ See that $H_k$ is a subgroup of $G.$ Now let $g:=\begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix} \in G$ and see that

$\displaystyle gH_kg^{-1}=\begin{pmatrix}1 & 2k\mathbb{Z} \\ 0 & 1\end{pmatrix}.$

So $gH_kg^{-1} \subset H_k$ but $gH_kg^{-1} \ne H. \ \Box$

## If x^n is quasi-regular, then so is x

Posted: February 28, 2020 in Elementary Algebra & Number Theory; Problems & Solutions, Rings and Modules
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Definition. Let $R$ be a ring. An element $x \in R$ is said to be right quasi-regular if there exists $y \in R$ such that $x+y=xy.$ Similarly, $x$ is said to be left quasi-regular if $x+y=yx$ for some $y \in R.$

Remark 1. If $R$ has $1,$ then $x+y=xy$ is equivalent to $(1-x)(1-y)=1.$ So, in this case, $x$ is right quasi-regular if and only if $1-x$ has a right inverse. Similarly, $x$ is left quasi-regular if and only if $1-x$ has a left inverse.

Remark 2. If $R$ is a ring and $x \in R$ is (right) quasi-regular, then $x^2$ is not necessarily (right) quasi-regular. For example, choose $R=\mathbb{Z}_3$ and $x=-1.$ The converse however is always true, as the following problem shows.

Problem. Let $R$ be a ring and let $x \in R.$ Show that if $x^n$ is (right) quasi-regular for some integer $n \ge 2,$ then $x$ is (right) quasi-regular too.

Solution. So there exists $y \in R$ such that

$x^n+y=x^ny.$

Let

$I:=\{xr-r: \ \ \ r \in R\}.$

See that $I$ is a right ideal of $R.$ To solve the problem, we only need to show that $x \in I.$ In $I$ we put $r=x^{k-1}y,$ where $k \ge 1$ is any integer, to get $x^ky-x^{k-1}y \in I$ (note that for $k=1,$ the term $x^{k-1}y$ simply means $y$ not $1 \cdot y,$ which is not necessarily meaningful since $R$ may not have $1$). Also, if in $I$ we put $r=x^{k-1},$ where $k \ge 2$ is any integer, we get $x^k-x^{k-1} \in I.$ Thus

\displaystyle \begin{aligned}x^{n-1}=x^n-(x^n-x^{n-1})=x^ny-y-(x^n-x^{n-1})=\sum_{k=1}^n(x^ky-x^{k-1}y)-(x^n-x^{n-1}) \in I\end{aligned}

and hence

$\displaystyle x=x^{n-1}-\sum_{k=2}^{n-1}(x^k-x^{k-1}) \in I. \ \Box$

## Aut(G) is non-trivial for groups of order > 2

Posted: February 28, 2020 in Elementary Algebra & Number Theory; Problems & Solutions, Groups and Fields
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Problem. Let $G$ be a group and suppose that $|G| > 2.$ Show that $|\text{Aut}(G)| > 1,$ i.e. $G$ has a non-trivial group automorphism.

Solution. We consider three cases.

Case 1: $G$ is non-abelian. In this case, let $Z(G)$ be the center of $G$ and choose $g \in G \setminus Z(G).$ Then the map $f: G \to G$ defined by $f(x)=gxg^{-1}$ is clearly an automorphism of $G$ and it is non-trivial because if $f(x)=x$ for all $x \in G,$ then $gx=xg$ for all $x \in G$ and so $g \in Z(G),$ which is not the case.

Case 2: $G$ is abelian and there exists $a \in G$ such that $a^2 \ne 1.$ In this case, the map $f: G \to G$ defined by $f(x)=x^{-1}$ is clearly an  automorphism of $G$ and it is non-trivial because $f(a) \ne a.$

Case 3: $G$ is abelian and $x^2 = 1$ for all $x \in G.$ In this case, looking at $G$ as an additive group, the condition $x^2=1$ becomes $2x=0$  and so we may consider $G$ as a vector space over the field $F:=\mathbb{Z}_2.$
Since $|G| > 2,$ we have $\dim_F G > 1.$ Let

$B:=\{x_i: \ \ i \in I\}$

be a basis for $G.$ Since $|B| \ge 2,$ we may choose two distinct elements $x_i, x_j \in B.$ Now the map $f: G \to G$ defined on $B$ by $f(x_i)=x_j, \ f(x_j)=f(x_i)$ and $f(x_k)=x_k, \ \ k \in I \setminus \{i,j\},$ and then  extended to the entire $G$ lineraly, is clearly a non-trivial automorphism of $G. \ \Box$

## Miklós Schweitzer 2019, Problem 2

Posted: December 29, 2019 in Noncommutative Ring Theory Notes, Simple Rings
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Problem (Miklós Schweitzer Competition, 2019). Let $R$ be a noncommutative finite ring with multiplicative identity element $1.$ Show that if the subring generated by $I \cup \{1\}$ is $R$ for each nonzero two-sided ideal $I,$ then $R$ is simple.

Solution. I show that, more generally, the result holds true for any noncommutative (left) artinian ring $R.$ Let $Z$ be the center of $R.$ For any two-sided ideal $I \ne (0)$ of $R,$ the subring $Z+I$ contains $I \cup \{1\}$ and so

$Z+I=R. \ \ \ \ \ \ \ \ \ (*)$

As a result, non-zero ideals of $R$ are noncommutative because $R$ is noncommutative.
We now show that $Z$ is a domain. Suppose, to the contrary, that $ab=0$ for some $a,b \in Z \setminus \{0\}.$ Then, by $(*),$ we have $Z+Ra=R$ and so $Zb=Rb$ implying that $Rb \ne (0)$ is a commutative ideal of $R,$ contradiction. So $Z$ is a domain.
Now let $J$ be the Jacobson radical of $R.$ We show that $J=(0).$ So suppose, to the contrary, that $J \ne (0).$ Since $R$ is artinian, $J$ is nilpotent. Thus there exists the smallest integer $n \ge 2$ such that $J^n=(0).$ Let $x \in J^{n-1} \setminus \{0\}.$ We have $Z+J=R,$ by $(*),$ and so $Rx=xR=Zx,$ i.e. $Rx \ne (0)$ is a commutative ideal of $R$ and that’s a contradiction. So $J=(0)$ and thus, by the Artin-Wedderburn’s theorem, $R$ is a finite direct product of some matrix rings over division rings

$R \cong \prod_{i=1}^kM_{n_i}(D_i).$

Let $F_i$ be the center of $D_i.$ Then $Z \cong \prod_{i=1}^k F_i$ which is possible only if $k=1$ because, as we showed, $Z$ is a domain. So $R \cong M_n(D)$ for some integer $n$ and some division ring $D$ and hence $R$ is simple. $\Box$

Remark. The above result is not true for commutative artinian rings. For example, let $R:=\mathbb{Z}/n\mathbb{Z},$ where $n > 1$ is not prime. Clearly $R$ is not simple and the only subring of $R$ that contains $1$ is $R$ itself. So the subring generated by $I \cup \{1\}$ is $R$ for any non-zero ideal $I$ of $R.$