Suppose (G, ∙ ) and (H, ∗ ) are groups. Take homomorphism φ : G → H. Suppose ∃g∈G : ∣g∣ = n, then ∣φ(g)∣ ≤ n. Does ∀g∈G, ∣g∣ = ∣φ(g)∣ ⇒ G ≅ H ?


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Question Number 189013 by Pengu last updated on 10/Mar/23

$Suppose (G, \cdot) and (H, *) are groups .$ $Take homomorphism ϕ : G \to H .$ $Suppose \exists g \in G : ∣ g ∣ = n, then ∣ ϕ (g) ∣ ⩽ n .$ $Does \forall g \in G, ∣ g ∣ = ∣ ϕ (g) ∣ \Rightarrow G ≅ H ?$
	Answered by aleks041103 last updated on 11/Mar/23

	$s u p p o s e ∣ ϕ (g) ∣= m > n .$ $\Rightarrow {(ϕ (g))}^{k} \neq {i d}_{H}, k = 1, 2, . . ., m - 1$ $B u t, s i n c e ϕ i s a h o m o m o r h i s m, t h e n$ ${(ϕ (g))}^{k} = ϕ (g^{k})$ $l e t k = n < m \Rightarrow {(ϕ (g))}^{n} = ϕ (g^{n}) = ϕ ({i d}_{G}) = {i d}_{H}$ $b u t s i n c e n < m, t h e n {(ϕ (g))}^{n} \neq {i d}_{H}$ $\Rightarrow {i d}_{H} \neq {i d}_{H} \Rightarrow c o n t r a d i c t i o n$ $\Rightarrow∣ ϕ (g) ∣⩽∣ g ∣$
	Commented by aleks041103 last updated on 11/Mar/23

	$N o t e :$ $ϕ : (G, \cdot) \to (H, ) i s a h o m o m o r h i s m$ $i f f$ $ϕ (g_{1} \cdot g_{2}) = ϕ (g_{1}) ϕ (g_{2})$ $\Rightarrow t h e f o l l o w i n g p r o p e r t i e s a r e f o l l o w i n g$ $ϕ (i d \cdot g) = ϕ (g) = ϕ (i d) * ϕ (g) = i d * ϕ (g)$ $\Rightarrow s i n c e i d i s u n i q u e t h e n$ $ϕ (i d) = i d$ $\Rightarrow ϕ (g^{- 1} \cdot g) = ϕ (g^{- 1}) * ϕ (g) = ϕ (i d) = i d$ $\Rightarrow ϕ (g^{- 1}) = {(ϕ (g))}^{- 1}$
	Answered by aleks041103 last updated on 11/Mar/23

	$F o r t h e s e c o n d q u e s t i o n :$ $i f ϕ \in H o m (G, H) i s s u r j e c t i v e$ $a n d \forall g \in G :∣ g ∣=∣ ϕ (g) ∣ t h e n G ≅ H .$ $s u p p o s e ϕ i s n t i n j e c t i v e$ $t h e n \exists g_{1}, g_{2} \in G a n d g_{1} \neq g_{2} s u c h t h a t$ $ϕ (g_{1}) = ϕ (g_{2})$ $\Rightarrow {(ϕ (g_{1}))}^{- 1} * ϕ (g_{1}) = {(ϕ (g_{1}))}^{- 1} * ϕ (g_{2})$ ${i d}_{H} = ϕ (g_{1}^{- 1} \cdot g_{2}) = ϕ (g_{3})$ $s i n c e ∣ g ∣=∣ ϕ (g) ∣\Rightarrow 1 =∣ {i d}_{H} ∣=∣ ϕ (g_{3}) ∣=∣ g_{3} ∣$ $\Rightarrow∣ g_{3} ∣= 1 \Rightarrow g_{3}^{1} = g_{3} = {i d}_{G}$ $\Rightarrow g_{1}^{- 1} \cdot g_{2} = {i d}_{G} \Rightarrow g_{1} = g_{2} \Rightarrow c o n t r a d i c t i o n$ $\Rightarrow ϕ i s i n j e c t i v e .$ $S i n c e ϕ i s a l s o s u r j e c t i v e \Rightarrow ϕ i s b i j e c t i v e$ $\Rightarrow ϕ i s b i j e c t i v e h o m o m o r p h i s m$ $\Rightarrow ϕ i s a i s o m o r p h i s m$ $\Rightarrow G ≅ H$ $i f G ≅ H t h e n \exists b i j . h o m o m o r h i s m ϕ : G \to H$ $t h e n i t i s o b v i o u s t h a t ∣ ϕ (g) ∣=∣ g ∣ f o r \forall g \in G$
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