# Prove that: a∈G⇒a=unvma \in G \Rightarrow a = u^n v^m with n,m∈{0,1,…,p−1}n, m \in \{0, 1, …, p-1\}

Is pp a prime number and G=(G,⋅)G = (G, \cdot) is not cyclic Group of order p2p^2:

Proof:

Let u,v∈Gu,v \in G that ord(u)=ord(v)=pord(u) = ord(v) = p

Let a∈G⇒a=unvma \in G \Rightarrow a = u^n v^m with n,m∈{0,1,…,p−1}n, m \in \{0, 1, …, p-1\}.

My Attachment:

u∈G⇒ord(u)⊆|G|⇒ord(u)⊆p2⇔ord(u)=pu \in G \Rightarrow ord(u) \subseteq |G| \Rightarrow ord(u) \subseteq p^2 \Leftrightarrow ord(u) = p
v∈G⇒ord(v)⊆|G|⇒ord(v)⊆p2⇔ord(v)=pv \in G \Rightarrow ord(v) \subseteq |G| \Rightarrow ord(v) \subseteq p^2 \Leftrightarrow ord(v) = p

ord(u)=ord(v)=p◻ord(u) = ord(v) = p \,\,\,\square

How can I prove that: a∈G⇒a=unvma \in G \Rightarrow a = u^n v^m with n,m∈{0,1,…,p−1}n, m \in \{0, 1, …, p-1\}.

a∈G⇒ord(a)⊆|G|⇒ord(a)⊆p2⇔ord(a)=p⇒ord(a)=ord(u)=ord(v)=pa \in G \Rightarrow ord(a) \subseteq |G| \Rightarrow ord(a) \subseteq p^2 \Leftrightarrow ord(a) = p \Rightarrow ord(a) = ord(u) = ord(v) = p

u∈G⇒ord(u)⊆|G|⇒|G|=ord(u)⋅n⇒u \in G \Rightarrow ord(u) \subseteq |G| \Rightarrow |G| = ord(u) \cdot n \Rightarrow

u|G|=e⇒uord(u)⋅n=uord(u)⋅un=e⋅un=un=eu^{|G|}=e \Rightarrow u^{ord(u) \cdot n} = u^{ord(u)} \cdot u^{n} = e \cdot u^{n} = u^{n} = e
v∈G⇒ord(v)⊆|G|⇒|G|=ord(v)⋅m⇒v \in G \Rightarrow ord(v) \subseteq |G| \Rightarrow |G| = ord(v) \cdot m \Rightarrow

v|G|=e⇒vord(v)⋅m=vord(v)⋅vm=e⋅vm=vm=ev^{|G|}=e \Rightarrow v^{ord(v) \cdot m} = v^{ord(v)} \cdot v^{m} = e \cdot v^{m} = v^{m} = e

vmun=ev^{m}u^{n}= e

How can I continue?

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– Charter
2 days ago

I need to prove thet a=unvma = u^n v^m
– Dario Gutierrez
2 days ago

1

But again, what’s your hypothesis? You wrote in your “proof” what you want to prove.
– Charter
2 days ago

You need to show that G≅A×BG \cong A \times B where |A|=|B|=p|A|=|B| = p. The rest follows almost immediately.
– Steven Gregory
2 days ago

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