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Suppose-G-and-H-are-groups-Take-homomorphism-G-H-Suppose-g-G-g-n-then-g-n-Does-g-G-g-g-G-H-

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Question Number 189013 by Pengu last updated on 10/Mar/23
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 ?
$$\mathrm{Suppose}\:\left({G},\:\centerdot\:\right)\:\mathrm{and}\:\left({H},\:\ast\:\right)\:\mathrm{are}\:\mathrm{groups}. \\ $$$$\mathrm{Take}\:\mathrm{homomorphism}\:\phi\::\:{G}\:\rightarrow\:{H}. \\ $$$$\mathrm{Suppose}\:\exists{g}\in{G}\::\:\mid{g}\mid\:=\:{n},\:\mathrm{then}\:\mid\phi\left({g}\right)\mid\:\leqslant\:{n}. \\ $$$$\: \\ $$$$\mathrm{Does}\:\forall{g}\in{G},\:\mid{g}\mid\:=\:\mid\phi\left({g}\right)\mid\:\Rightarrow\:{G}\:\cong\:{H}\:? \\ $$
Answered by aleks041103 last updated on 11/Mar/23
suppose ∣φ(g)∣=m>n. ⇒(φ(g))^k ≠id_H , k=1,2,...,m−1 But, since φ is a homomorhism, then (φ(g))^k =φ(g^k ) let k=n<m⇒(φ(g))^n =φ(g^n )=φ(id_G )=id_H but since n<m, then (φ(g))^n ≠id_H ⇒id_H ≠id_H ⇒contradiction ⇒∣φ(g)∣≤∣g∣
$${suppose}\:\mid\phi\left({g}\right)\mid={m}>{n}. \\ $$$$\Rightarrow\left(\phi\left({g}\right)\right)^{{k}} \neq{id}_{{H}} ,\:{k}=\mathrm{1},\mathrm{2},…,{m}−\mathrm{1} \\ $$$${But},\:{since}\:\phi\:{is}\:{a}\:{homomorhism},\:{then} \\ $$$$\left(\phi\left({g}\right)\right)^{{k}} =\phi\left({g}^{{k}} \right) \\ $$$${let}\:{k}={n}<{m}\Rightarrow\left(\phi\left({g}\right)\right)^{{n}} =\phi\left({g}^{{n}} \right)=\phi\left({id}_{{G}} \right)={id}_{{H}} \\ $$$${but}\:{since}\:{n}<{m},\:{then}\:\left(\phi\left({g}\right)\right)^{{n}} \neq{id}_{{H}} \\ $$$$\Rightarrow{id}_{{H}} \neq{id}_{{H}} \Rightarrow{contradiction} \\ $$$$\Rightarrow\mid\phi\left({g}\right)\mid\leqslant\mid{g}\mid \\ $$
Commented by aleks041103 last updated on 11/Mar/23
Note: φ:(G,∙)→(H,∗) is a homomorhism iff φ(g_1 ∙g_2 )=φ(g_1 )∗φ(g_2 ) ⇒ the following properties are following φ(id∙g)=φ(g)=φ(id)∗φ(g)=id∗φ(g) ⇒since id is unique then φ(id)=id ⇒φ(g^(−1) ∙g)=φ(g^(−1) )∗φ(g)=φ(id)=id ⇒φ(g^(−1) )=(φ(g))^(−1)
$${Note}: \\ $$$$\phi:\left({G},\centerdot\right)\rightarrow\left({H},\ast\right)\:{is}\:{a}\:{homomorhism} \\ $$$${iff} \\ $$$$\phi\left({g}_{\mathrm{1}} \centerdot{g}_{\mathrm{2}} \right)=\phi\left({g}_{\mathrm{1}} \right)\ast\phi\left({g}_{\mathrm{2}} \right) \\ $$$$\Rightarrow\:{the}\:{following}\:{properties}\:{are}\:{following} \\ $$$$ \\ $$$$\phi\left({id}\centerdot{g}\right)=\phi\left({g}\right)=\phi\left({id}\right)\ast\phi\left({g}\right)={id}\ast\phi\left({g}\right) \\ $$$$\Rightarrow{since}\:{id}\:{is}\:{unique}\:{then} \\ $$$$\phi\left({id}\right)={id} \\ $$$$ \\ $$$$\Rightarrow\phi\left({g}^{−\mathrm{1}} \centerdot{g}\right)=\phi\left({g}^{−\mathrm{1}} \right)\ast\phi\left({g}\right)=\phi\left({id}\right)={id} \\ $$$$\Rightarrow\phi\left({g}^{−\mathrm{1}} \right)=\left(\phi\left({g}\right)\right)^{−\mathrm{1}} \\ $$
Answered by aleks041103 last updated on 11/Mar/23
For the second question: if φ∈Hom(G,H) is surjective and ∀g∈G:∣g∣=∣φ(g)∣ then G≅H. suppose φ isnt injective then ∃g_1 ,g_2 ∈G and g_1 ≠g_2 such that φ(g_1 )=φ(g_2 ) ⇒(φ(g_1 ))^(−1) ∗φ(g_1 )=(φ(g_1 ))^(−1) ∗φ(g_2 ) id_H =φ(g_1 ^(−1) ∙g_2 )=φ(g_3 ) since ∣g∣=∣φ(g)∣⇒1=∣id_H ∣=∣φ(g_3 )∣=∣g_3 ∣ ⇒∣g_3 ∣=1⇒g_3 ^1 =g_3 =id_G ⇒g_1 ^(−1) ∙g_2 =id_G ⇒g_1 =g_2 ⇒contradiction ⇒φ is injective. Since φ is also surjective⇒φ is bijective ⇒φ is bijective homomorphism ⇒φ is a isomorphism ⇒G≅H if G≅H then ∃bij. homomorhism φ:G→H then it is obvious that ∣φ(g)∣=∣g∣ for ∀g∈G
$${For}\:{the}\:{second}\:{question}: \\ $$$${if}\:\phi\in{Hom}\left({G},{H}\right)\:{is}\:{surjective} \\ $$$$\:{and}\:\forall{g}\in{G}:\mid{g}\mid=\mid\phi\left({g}\right)\mid\:{then}\:{G}\cong{H}. \\ $$$$ \\ $$$${suppose}\:\phi\:{isnt}\:{injective} \\ $$$${then}\:\:\exists{g}_{\mathrm{1}} ,{g}_{\mathrm{2}} \in{G}\:{and}\:{g}_{\mathrm{1}} \neq{g}_{\mathrm{2}} \:{such}\:{that} \\ $$$$\phi\left({g}_{\mathrm{1}} \right)=\phi\left({g}_{\mathrm{2}} \right) \\ $$$$\Rightarrow\left(\phi\left({g}_{\mathrm{1}} \right)\right)^{−\mathrm{1}} \ast\phi\left({g}_{\mathrm{1}} \right)=\left(\phi\left({g}_{\mathrm{1}} \right)\right)^{−\mathrm{1}} \ast\phi\left({g}_{\mathrm{2}} \right) \\ $$$${id}_{{H}} =\phi\left({g}_{\mathrm{1}} ^{−\mathrm{1}} \centerdot{g}_{\mathrm{2}} \right)=\phi\left({g}_{\mathrm{3}} \right) \\ $$$${since}\:\mid{g}\mid=\mid\phi\left({g}\right)\mid\Rightarrow\mathrm{1}=\mid{id}_{{H}} \mid=\mid\phi\left({g}_{\mathrm{3}} \right)\mid=\mid{g}_{\mathrm{3}} \mid \\ $$$$\Rightarrow\mid{g}_{\mathrm{3}} \mid=\mathrm{1}\Rightarrow{g}_{\mathrm{3}} ^{\mathrm{1}} ={g}_{\mathrm{3}} ={id}_{{G}} \\ $$$$\Rightarrow{g}_{\mathrm{1}} ^{−\mathrm{1}} \centerdot{g}_{\mathrm{2}} ={id}_{{G}} \Rightarrow{g}_{\mathrm{1}} ={g}_{\mathrm{2}} \Rightarrow{contradiction} \\ $$$$\Rightarrow\phi\:{is}\:{injective}. \\ $$$${Since}\:\phi\:{is}\:{also}\:{surjective}\Rightarrow\phi\:{is}\:{bijective} \\ $$$$\Rightarrow\phi\:{is}\:{bijective}\:{homomorphism} \\ $$$$\Rightarrow\phi\:{is}\:{a}\:{isomorphism} \\ $$$$\Rightarrow{G}\cong{H} \\ $$$$ \\ $$$${if}\:{G}\cong{H}\:{then}\:\exists{bij}.\:{homomorhism}\:\phi:{G}\rightarrow{H} \\ $$$${then}\:{it}\:{is}\:{obvious}\:{that}\:\mid\phi\left({g}\right)\mid=\mid{g}\mid\:{for}\:\forall{g}\in{G} \\ $$$$ \\ $$

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