Datasets:

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aa_test

like 0

import Mathlib example { G : Type* } [ Group G ] { M N : Subgroup G } ( hM : M.Normal ) ( hN : N.Normal ) ( h : M ⊓ N = ⊥ ) { a : G } ( ha : a ∈ M ) { b : G } ( hb : b ∈ N ) : a * b = b * a := by have : (h : b * a * b⁻¹ * a⁻¹ = 1) → a * b = b * a := by intro h calc _ = 1 * a * b := by group _ = (b * a * b⁻¹ * a⁻¹) * a * b := by rw [← h] _ = _ := by group have α : b * a * b⁻¹ * a⁻¹ ∈ M := M.mul_mem (hM.conj_mem a ha b) (M.inv_mem ha) have β : b * (a * b⁻¹ * a⁻¹) ∈ N := N.mul_mem hb (hN.conj_mem b⁻¹ (N.inv_mem hb) a) repeat rw [← mul_assoc] at β have γ : b * a * b⁻¹ * a⁻¹ ∈ M ⊓ N := ⟨α,β⟩ rw [h] at γ have δ : b * a * b⁻¹ * a⁻¹ = 1 := by exact γ exact this δ

Let $M$ and $N$ be normal subgroups of the group $G$. If $M \cap N = 1$, then for any $a \in M$ and $b \in N$, we have $ab = ba$.

import Mathlib

import Mathlib

{"env": 0}

import Mathlib --1203 example {G : Type*} [Group G]: ∃! (le : G), ∃! (re : G), ∀ (g : G), le * g = g ∧ g * re = g := by use 1 dsimp constructor · use 1 dsimp constructor · group simp only [and_self, implies_true] · intro y H have : 1 * y = 1 := (H 1).2 calc _ = 1 * y := by group _ = _ := by rw[this] · intro y h rcases h with ⟨re,h1,_⟩ have l : y * 1 = 1 := (h1 1).1 rw[mul_one] at l rw[l]

Prove that a group have only one left identity element and only one right identity element.

import Mathlib example {G : Type*} [Group G]: ∃! (le : G), ∃! (re : G), ∀ (g : G), le * g = g ∧ g * re = g := by sorry

import Mathlib

{"env": 0}

import Mathlib def Func (S : Type*) := S → ℝ instance (S : Type*) : Add (Func S) where add := λ f g => λ x => f x + g x instance (S : Type*) : Neg (Func S) where neg := λ f => λ x => - f x instance (S : Type*) : Zero (Func S) where zero := λ _ => 0 instance (S : Type*) : AddGroup (Func S) := by apply AddGroup.ofLeftAxioms · intro f g h funext x exact add_assoc (f x) (g x) (h x) · intro f funext x exact zero_add (f x) · intro f funext x exact add_left_neg (f x)

Consider a set $S$, let $G:=\{f:S\to\mathbb R\}$ to be all maps from $S$ to $\mathbb R$. Then $G$ is a group under function addition.

import Mathlib.Tactic import Mathlib.Algebra.Group.MinimalAxioms def Func (S : Type*) := S → ℝ instance (S : Type*) : Add (Func S) where add := sorry instance (S : Type*) : Neg (Func S) where neg := sorry instance (S : Type*) : Zero (Func S) where zero := sorry noncomputable instance (S : Type*) : AddGroup (Func S) := by apply AddGroup.ofLeftAxioms · sorry · sorry · sorry

import Mathlib

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import Mathlib -- 著者：秦宇轩 import Mathlib.Algebra.Group.Basic import Mathlib.Tactic -- Let `G` be a group. -- Prove that `(a ^ {-1}) ^ {-1} = a` for any element `a ∈ G`. example {G : Type*} [Group G] {a :G} : (a⁻¹)⁻¹ = a := by group

Let $G$ be a group. Prove that $\left(a^{-1}\right)^{-1}=a$ for any element $a\in G$.

import Mathlib example {G : Type*}[Group G]{a :G}: (a⁻¹)⁻¹=a:=by sorry

import Mathlib

{"env": 0}

Proof 1. For the first theorem, we use the theorem that m is larger or equal than 2 if m isn't 1 and isn't 0. 2. And then, the conclusion becomes obvious, for it is exactly h0 and h1. 3. For the example, we use the theorem that that there is a prime which n could be divided by if n is not 1. 4. And then use n is larger or equal to 2 to get n is not 1. This finishes the proof.

import Mathlib theorem two_le {m : ℕ} (h0 : m ≠ 0) (h1 : m ≠ 1) : 2 ≤ m := by refine (Nat.two_le_iff m).mpr ?_ --For the first theorem, we use the theorem that m is larger or equal than 2 if m isn't 1 and isn't 0. constructor · exact h0 · exact h1 --And then, the conclusion becomes obvious, for it is exactly h0 and h1. example {n : Nat} (h : 2 ≤ n) : ∃ p : Nat, p.Prime ∧ p ∣ n := by refine Nat.exists_prime_and_dvd ?hn --For the example, we use the theorem that that there is a prime which n could be divided by if n is not 1. exact Ne.symm (Nat.ne_of_lt h) --And then use n is larger or equal to 2 to get n is not 1. This finishes the proof.

Given an nature number $n$ which is not less than 2, prove either it is a prime or it can be divided by a prime nature number.

import Mathlib theorem two_le {m : ℕ} (h0 : m ≠ 0) (h1 : m ≠ 1) : 2 ≤ m := sorry example {n : Nat} (h : 2 ≤ n) : ∃ p : Nat, p.Prime ∧ p ∣ n := sorry

import Mathlib

{"env": 0}

import Mathlib example {G : Type*} [Group G]: ∀ (g : G), ∃! (y : G), g * y = 1 := by intro g use g⁻¹ constructor · apply mul_right_inv intro y h rw [← mul_right_inv g] at h apply mul_left_cancel at h exact h

Prove that in a group, every element has exactly one inverse.

import Mathlib example {G : Type*} [Group G]: ∀ (g : G), ∃! (y : G), g * y = 1 := by sorry

import Mathlib

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Proof that if every element in a group \(G\) satisfies \(x \cdot x = 1\) for all \(x \in G\), then \(G\) is a commutative grouTo show that G is a commutative group, apply the constructor for CommGroup a * b = a * b * (a * b) * (a * b)⁻¹ = 1 * (a * b)⁻¹ = b⁻¹ * (b⁻¹⁻¹ * b) * a = b⁻¹ * (b⁻¹⁻¹ * b) * (a * a * a⁻¹) = 1 * (a * b)⁻¹ = b⁻¹ * 1 * (a * a * a⁻¹) = b⁻¹ * 1 * (1 * a⁻¹), the goal is changed to 1 * (a * b)⁻¹ = b⁻¹ * 1 * (1 * a⁻¹) Conclude that G is a commutative group by using the group properties

import Mathlib.Tactic example {G : Type*} [Group G] (h: ∀ (x : G), x * x = 1) : CommGroup G := by -- To show that G is a commutative group, apply the constructor for CommGroup apply CommGroup.mk intro a b -- a * b = a * b * (a * b) * (a * b)⁻¹ = 1 * (a * b)⁻¹ rw [← mul_inv_cancel_right (a * b) (a * b), h] -- = b⁻¹ * (b⁻¹⁻¹ * b) * a = b⁻¹ * (b⁻¹⁻¹ * b) * (a * a * a⁻¹) nth_rw 2 [← mul_inv_cancel_left b⁻¹ b, ← mul_inv_cancel_right a a] -- = 1 * (a * b)⁻¹ = b⁻¹ * 1 * (a * a * a⁻¹) = b⁻¹ * 1 * (1 * a⁻¹), the goal is changed to 1 * (a * b)⁻¹ = b⁻¹ * 1 * (1 * a⁻¹) rw [inv_inv, h, h] -- Conclude that G is a commutative group by using the group properties group

Show that every group $G$ with identity $e$ and such that $x * x=e$ for all $x \in G$ is Abelian.

import Mathlib example {G : Type*} [Group G] (h: ∀ (x : G), x * x = 1) : CommGroup G := by sorry

import Mathlib

{"env": 0}

import Mathlib example {G : Type*} [Group G] (a b : G) (h : a * b = b * a⁻¹) : b * a = a⁻¹ * b := by have h₀ : a⁻¹ * (a * b) * a = a⁻¹ * (b * a⁻¹) * a := by rw [h] calc _ = 1 * (b * a) := by rw [one_mul] _ = a⁻¹ * a * (b * a) := by rw [← mul_left_inv] _ = a⁻¹ * (a * b) * a := by rw [mul_assoc a⁻¹, ← mul_assoc a, ← mul_assoc] _ = a⁻¹ * (b * a⁻¹) * a := by rw [h₀] _ = a⁻¹ * b * (a⁻¹ * a) := by rw [mul_assoc, mul_assoc, ← mul_assoc] _ = a⁻¹ * b * 1 := by rw [mul_left_inv] _ = _ := by rw [mul_one]

Suppose that $G$ is a group and $a, b \in G$ satisfy $a * b=b * a^{-1}$. Prove that $b * a=a^{-1} * b$.

import Mathlib example {G : Type*} [Group G] (a b : G) (h : a * b = b * a⁻¹) : b * a = a⁻¹ * b := sorry

import Mathlib

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import Mathlib example {G : Type*} [Group G] (h: ∀ (x : G), x * x = 1) : CommGroup G := { mul_comm := by intro a b have h1 : (a * b) * (a * b) = 1 := by rw [h (a * b)] apply (mul_left_inj (a * b)⁻¹).mpr at h1 rw [mul_assoc, mul_right_inv (a * b), mul_one, one_mul] at h1 rw [h1] apply (mul_left_inj (a * b)).mp group rw [mul_assoc b a a, h] group rw [h] }

Show that every group $G$ with identity $e$ and such that $x * x=e$ for all $x \in G$ is Abelian.

import Mathlib example {G : Type*} [Group G] (h: ∀ (x : G), x * x = 1) : CommGroup G := by sorry

import Mathlib

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import Mathlib example {G : Type*} [Group G] {x y : G} (h : (x * y) * (x * y) = (x * x) * (y * y)): x * y = y * x := by rw[←mul_assoc, ←mul_assoc] at h have h1 : (x * y * x) * y * y⁻¹ = (x * x * y) * y * y⁻¹ := by rw[h] simp only [mul_inv_cancel_right] at h1 have h2 : x⁻¹ * (x * y * x) = x⁻¹ * (x * x * y) := by rw[h1,← mul_assoc, ← mul_assoc] rw[← mul_assoc,← mul_assoc,← mul_assoc,← mul_assoc] at h2 simp only [mul_left_inv, one_mul] at h2 exact id (Eq.symm h2)

Suppose that $G$ is a group with operation $\circ$; suppose that $x, y \in G$. Show that if $$ (x \circ y) \circ(x \circ y)=(x \circ x) \circ(y \circ y), $$ then $x \circ y=y \circ x$.

import Mathlib example {G : Type*} [Group G] {x y : G} (h : (x * y) * (x * y) = (x * x) * (y * y)): x * y = y * x := by sorry

import Mathlib

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import Mathlib example {G : Type*} [Group G] (a b c : G) : ∃! x : G, a * x * b = c := by use a⁻¹ * c * b⁻¹ constructor . group . intro x intro h have l : a⁻¹ * (a * x * b) * b⁻¹ = a⁻¹ * c * b⁻¹ := by rw [h] group at l group exact l

Suppose that $G$ is a group and $a, b, c\in G$. Prove that $a * x * b=c$ has a unique solution.

import Mathlib example {G : Type*} [Group G] (a b c : G) : ∃! x : G, a * x * b = c := sorry

import Mathlib

{"env": 0}

import Mathlib example (n : ℕ) (k j : Fin n) : MulAction.stabilizer (Equiv.Perm (Fin n)) k ≃ MulAction.stabilizer (Equiv.Perm (Fin n)) j where toFun := by intro α use (Equiv.swap k j) * α * (Equiv.swap k j) sorry invFun := sorry left_inv := sorry right_inv := sorry

16. Let $n$ be positive integer, and $k, j$ fixed integers with $1 \leq k, j \leq n$. Define the function $\Phi: \operatorname{Stab}(k) \rightarrow \operatorname{Stab}(j)$ on $\alpha \in \operatorname{Stab}(k)$ by setting $\Phi(\alpha)=(k j) \alpha(k j)$ (where $(k j)$ is the 2-cycle switching $k$ and $j$ ). Prove that $\Phi$ is a group isomorphism. Warning: An important verification is to check that $\Phi(\alpha) \in \operatorname{Stab}(j)$.

import Mathlib

import Mathlib

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import Mathlib example {G : Type*} [Group G] {x y : G} (h : (x * y) * (x * y) = (x * x) * (y * y)): x * y = y * x := by have lemma1 : x * y * x = x * x * y := by calc _ = (x * y) * (x * y) * y⁻¹ := by group _ = (x * x) * (y * y) * y⁻¹ := by rw [h] _ = _ := by group calc _ = x⁻¹ * (x * x * y) := by group _ = x⁻¹ * (x * y * x) := by rw[lemma1] _ = y * x := by group

Suppose that $G$ is a group with operation $\circ$; suppose that $x, y \in G$. Show that if $$ (x \circ y) \circ(x \circ y)=(x \circ x) \circ(y \circ y), $$ then $x \circ y=y \circ x$.

import Mathlib example {G : Type*} [Group G] {x y : G} (h : (x * y) * (x * y) = (x * x) * (y * y)): x * y = y * x := by sorry

import Mathlib

{"env": 0}

1、证明x的存在性 2、证明x的唯一性

import Mathlib section Basic_Calculations example {G : Type*} [Group G] (a b c : G) : ∃! x : G, a * x * b = c := by ·use a⁻¹ * c * b⁻¹ group rw[true_and] intro y h calc _ = a⁻¹ *(a*y*b)* b⁻¹ := by group _ = _ := by simp[h]

Suppose that $G$ is a group and $a, b,a_1,a_2 \in G$. Prove that $(bab^{-1})^{-1}=ba^{-1}b^{-1}$ and $(ba_1b^{-1})(ba_2b^{-1})=ba_1a_2b^{-1}$.

import Mathlib example {G : Type*} [Group G] (a b a₁ a₂ : G) : (b * a * b⁻¹)⁻¹ = b * a⁻¹ * b⁻¹ ∧ (b * a₁ * b⁻¹) * (b * a₂ * b⁻¹) = b * a₁ * a₂ * b⁻¹ := sorry

import Mathlib

{"env": 0}

We just enumerate all the conditions and use ‘decide’ to calculate the results.

import Mathlib.Tactic example : ∀ x : ZMod 7, 3 * x = 2 ↔ x = 3 := by intro x constructor -- We start by simplifying the negation of the existential quantifier. · by_contra h0 push_neg at h0 have h1 : 3 * x = 2 := h0.1 have h2 : x ≠ 3 := h0.2 absurd h1 -- We use pattern matching to check the equation for each possible value of x. match x with | 0 => decide | 1 => decide | 2 => decide | 4 => decide | 5 => decide | 6 => decide · intro h1 rw [h1]; decide example : ∀ x : ZMod 23, 3 * x = 2 ↔ x = 16 := by intro x constructor -- We start by simplifying the negation of the existential quantifier. · by_contra h0 push_neg at h0 have h1 : 3 * x = 2 := h0.1 have h2 : x ≠ 16 := h0.2 absurd h1 by_cases l0 : x = 16 exact fun a ↦ h2 l0 -- As for the conditions that x ≠ 16, we can easily eliminate it. fin_cases x <;> try decide have := h0.2 contradiction · intro h rw [h]; decide

Solve the equation $3 x=2$ in the field $\mathbb{Z}/7\mathbb{Z}$ and in the field $\mathbb{Z}/23\mathbb{Z}$.

import Mathlib

import Mathlib

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import Mathlib -- 著者：秦宇轩 import Mathlib.Algebra.Group.Basic import Mathlib.Tactic import Init.Prelude example {G : Type*} [Group G] (a : G) (k : ℕ) (h : Odd k) (pow_one : a ^ k = 1): Odd (orderOf a) := by have : (orderOf a) ∣ k := by exact orderOf_dvd_of_pow_eq_one pow_one exact Odd.of_dvd_nat h this

Let $a$ be an element of a group $G$. If $a^{k}=e$ where $k$ is odd, then the order of $a$ is odd.

import Mathlib example {G : Type*} [Group G] (a : G) (k : ℕ) (h : Odd k) (pow_one : a ^ k = 1): Odd (orderOf a) := by sorry

import Mathlib

{"env": 0}

1. We define a structure two_div_sum with two integer $x,y$ components and a proof $h$ that $2 ∣ x + y$. 2. The $Add$ on two_div_sum is defined as $(a+b).x=a.x+b.x, (a+b).y=a.y+b.y$ and $2 ∣ (a+b).x + (a+b).y$. 3. The $Neg$ on two_div_sum is defined as $(-a).x=-a.x, (-a).y=-a.y$ and $2 ∣ (-a).x + (-a).y$. 4. The $Zero$ on two_div_sum is defined as $0.x=0, 0.y=0$ and $2 ∣ 0.x + 0.y$. 5. We need prove that two_div_sum forms an additive group. Just need to prove the three axioms of additive group: associativity, identity element and inverse element. We prove these instance in its components, then is trival.

import Mathlib -- Define a structure 'two_div_sum' with two integer $x,y$ components and a proof $h$ that $2 ∣ x + y$ @[ext] structure two_div_sum where x : ℤ y : ℤ h : 2 ∣ x + y -- Define an 'Add' instance for 'two_div_sum' using the properties of integers instance : Add two_div_sum where add := by -- Introduce two elements a and b of the structure intro a b -- Define the $sum.x$ as $a.x + b.x$ use a.x + b.x -- Define the $sum.y$ as $a.y + b.y$ use a.y + b.y -- Prove that $2 ∣ sum.x + sum.y$ have : a.x + b.x + (a.y + b.y) = a.x + a.y +(b.x + b.y) := by ring rw [this] apply dvd_add · exact a.h · exact b.h -- Define a 'Neg' instance for 'two_div_sum' using the properties of integers instance : Neg two_div_sum where neg := by -- Introduce an element a of the structure intro a -- Define the $neg.x$ as $-a.x$ use -a.x -- Define the $neg.y$ as $-a.y$ use -a.y -- Prove that $2 ∣ neg.x + neg.y$ rw [← neg_add] apply Dvd.dvd.neg_right exact a.h -- Define a 'Zero' instance for 'two_div_sum' using the properties of integers instance : Zero two_div_sum where -- Define the zero element of the structure zero := by -- Define the $zero.x$ as $0$ use 0 -- Define the $zero.y$ as $0$ use 0 -- Prove that $2 ∣ zero.x + zero.y$ simp only [add_zero, dvd_zero] -- Prove that 'two_div_sum' forms an additive group instance : AddGroup two_div_sum := by apply AddGroup.ofLeftAxioms -- Prove that the addition operation is associative · intro a b c --Prove the equation in its components ext --Use the ring tactic to prove the equation · calc _ = a.x + b.x + c.x := by rfl _ = a.x + (b.x + c.x) := by ring _ = _ := by rfl · calc _ = a.y + b.y + c.y := by rfl _ = a.y + (b.y + c.y) := by ring _ = _ := by rfl -- Prove that the zero element is the identity element · intro a --Prove the equation in its components ext --Use the ring tactic to prove the equation · calc _ = 0 + a.x := by rfl _ = a.x := by ring _ = _ := by rfl · calc _ = 0 + a.y := by rfl _ = a.y := by ring _ = _ := by rfl -- Prove that the negation operation is the inverse operation · intro a ----Use the ring tactic to prove the equation ext --Use the ring tactic to prove the equation · calc _ = -a.x + a.x := by rfl _ = 0 := by ring _ = _ := by rfl · calc _ = -a.y + a.y := by rfl _ = 0 := by ring _ = _ := by rfl

Prove that the elements $\{(x,y):x,y\in\mathbb Z, 2\mid x+y\}$ under elementwise addition is a group.

import Mathlib.Tactic import Mathlib.Algebra.Group.MinimalAxioms structure two_div_sum where x : ℤ y : ℤ h : 2 ∣ x + y instance : Add two_div_sum where add := sorry instance : Neg two_div_sum where neg := sorry instance : Zero two_div_sum where zero := sorry instance : AddGroup two_div_sum := by apply AddGroup.ofLeftAxioms · sorry · sorry · sorry

import Mathlib

{"env": 0}

1. first we show that order of g * a * g⁻¹ is 2 2. by calculation, (g * a * g⁻¹) ^ 2 = g * (a * a) * g⁻¹ = 1 3. so the order of g * a * g⁻¹ can only be 1 or 2, we show it is not 1. if so a = 1, which contradicts the assumption that order of a is 2 4. since a is the unique element with order 2, we have g * a * g⁻¹ = a, i.e. g * a = a * g, which is what we want to prove.

import Mathlib.Tactic example {G : Type*} [Group G] {a : G} (h₁: orderOf a = 2) (h₂: ∃! (x : G), orderOf x = 2) : ∀ x : G, a * x = x * a := by intro g rw [ExistsUnique] at h₂ rcases h₂ with ⟨x, ⟨ _ , hx2⟩⟩ have haa: a * a = 1 := by calc a * a = a ^ 2 := by exact Eq.symm (pow_two a) _ = 1 := by rw [←h₁ ,pow_orderOf_eq_one a] -- first we show that order of g * a * g⁻¹ is 2 have ho2: orderOf (g * a * g⁻¹) = 2 := by rw [orderOf_eq_iff ] constructor -- by calculation, (g * a * g⁻¹) ^ 2 = g * (a * a) * g⁻¹ = 1 . calc (g * a * g⁻¹) ^ 2 = (g * a * g⁻¹) * (g * a * g⁻¹) := by exact pow_two (g * a * g⁻¹) _ = g * (a * a) * g⁻¹ := by group _ = g * 1 * g⁻¹ := by rw [haa] _ = 1 := by group -- so the order of g * a * g⁻¹ can only be 1 or 2, we show it is not 1. if so a = 1, which contradicts the assumption that order of a is 2 . intro m hm2 h0m hf have hm: m = 1 := by exact Nat.eq_of_le_of_lt_succ h0m hm2 rw [hm] at hf have ha1: a = 1 := by calc a = g⁻¹ * (g * a * g⁻¹) ^ 1 * g := by group _ = g⁻¹ * 1 * g := by rw [hf] _ = 1 := by group have hao1 : orderOf a = 1 := by exact orderOf_eq_one_iff.mpr ha1 have hn12: 1 ≠ 2 := by linarith apply hn12 calc 1 = orderOf a := by rw [hao1] _ = 2 := by rw [h₁] simp only [Nat.ofNat_pos] -- since a is the unique element with order 2, we have g * a * g⁻¹ = a, i.e. g * a = a * g, which is what we want to prove. suffices g * a * g⁻¹ = x ∧ a = x by calc a * g = x * g := by rw [this.2] _ = g * a * g⁻¹ * g := by rw [this.1] _ = g * a := by group constructor . exact hx2 (g * a * g⁻¹) ho2 . exact hx2 a h₁

Let $G$ be a group and suppose $a \in G$ generates a cyclic subgroup of order 2 and is the unique such element. Show that $a x=x a$ for all $x \in G$. [Hint: Consider $\left(x a x^{-1}\right)^{2}$.]

import Mathlib example {G : Type*} [Group G] {a : G} (h₁: orderOf a = 2) (h₂: ∃! (x : G), orderOf x = 2) : ∀ x : G, a * x = x * a := by sorry

import Mathlib

{"env": 0}

import Mathlib instance {G : Type*} [Group G] (h : IsCyclic G) : CommGroup G := by apply CommGroup.mk intro a b rcases h with ⟨t, ht⟩ rcases ht a with ⟨na, hna⟩ simp at hna rcases ht b with ⟨nb, hnb⟩ simp at hnb rw [← hna, ← hnb] group

Every cyclic group is a abelian group.

import Mathlib instance : sorry := sorry

import Mathlib

{"env": 0}

import Mathlib example {G : Type*} [Group G] (a b : G) (h : a * b = b * a⁻¹) : b * a = a⁻¹ * b := by have h₁ : a * b * a = b := by rw [h, mul_assoc, mul_left_inv, mul_one] calc _ = a⁻¹ * a * b * a := by rw [mul_assoc, mul_left_inv, one_mul] _ = _ := by rw [mul_assoc,mul_assoc] nth_rw 2 [← mul_assoc] rw [h₁]

Suppose that $G$ is a group and $a, b \in G$ satisfy $a * b=b * a^{-1}$. Prove that $b * a=a^{-1} * b$.

import Mathlib example {G : Type*} [Group G] (a b : G) (h : a * b = b * a⁻¹) : b * a = a⁻¹ * b := sorry

import Mathlib

{"messages": [{"severity": "warning", "pos": {"line": 5, "column": 22}, "endPos": {"line": 5, "column": 34}, "data": "`mul_left_inv` has been deprecated, use `inv_mul_cancel` instead"}, {"severity": "warning", "pos": {"line": 8, "column": 21}, "endPos": {"line": 8, "column": 33}, "data": "`mul_left_inv` has been deprecated, use `inv_mul_cancel` instead"}], "env": 0}