Abstract-Algebra-1-Group-Homomorphisms, guaranteed and verified 100% Pass

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Group Homomorphisms
Recall that an isomorphism of a group 𝐺 with a group 𝐺 ′ is a 1-1 function
mapping 𝐺 onto 𝐺 ′ such that:

𝜙(𝑥𝑦) = 𝜙(𝑥)𝜙(𝑦) for all 𝑥, 𝑦 ∈ 𝐺.
Isomorphic groups have the same group structure.


Def. A map 𝜙: 𝐺 → 𝐺 ′ is a group homomorphism if

𝜙(𝑥𝑦) = 𝜙(𝑥)𝜙(𝑦) for all 𝑥, 𝑦 ∈ 𝐺.
An isomorphism is a type of homomorphism that is 1-1 and onto.


For any groups 𝐺, 𝐺 ′ there is always at least one homomorphism, 𝜙: 𝐺 → 𝐺 ′ ,
given by 𝜙(𝑔) = 𝑒 ′ for all 𝑔 ∈ 𝐺. This is called the trivial homomorphism.
However, this is not a very useful homomorphism because no information about
the group structures of 𝐺 and 𝐺 ′ can be gained from this.



Ex. Let 𝜙: 𝐺1 → 𝐺2 be a homomorphism of 𝐺1 into 𝐺2 . Show that If 𝐺1 is

abelian and 𝜙 is onto then 𝐺2 must be abelian. However, if 𝜙 is not onto

then 𝐺2 need not be abelian.



To show 𝐺2 is abelian we must show given any 𝑎2 , 𝑏2 ∈ 𝐺2 that
𝑎2 𝑏2 = 𝑏2 𝑎2 .
Since 𝜙 is onto, we know there is at least one 𝑎1 ∈ 𝐺1 and at least one

𝑏1 ∈ 𝐺1 such that 𝜙(𝑎1 ) = 𝑎2 and 𝜙(𝑏1 ) = 𝑏2 .

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Since 𝐺1 is abelian 𝑎1 𝑏1 = 𝑏1 𝑎1 . So we know:

𝑎2 𝑏2 = 𝜙(𝑎1 )𝜙(𝑏1 ) = 𝜙(𝑎1 𝑏1 ) = 𝜙(𝑏1 𝑎1 )
= 𝜙(𝑏1 )𝜙(𝑎1 ) = 𝑏2 𝑎2
Thus, 𝐺2 is abelian.



Notice that if 𝜙 is not onto then we can have the trivial homomorphism:

𝜙: ℤ6 → 𝑆3 , 𝜙(𝑘) = 𝜌0 = 𝑖𝑑𝑒𝑛𝑡𝑖𝑡𝑦.
Thus 𝐺1 = ℤ6 is abelian but 𝐺2 = 𝑆3 is non-abelian.




Ex. Let 𝑟 ∈ ℤ. Consider two mappings from ℤ, + to ℤ,+ :

𝜙1 : ℤ → ℤ; 𝜙1 (𝑛) = 𝑟𝑛 for all 𝑛 ∈ ℤ
𝜙2 : ℤ → ℤ; 𝜙2 (𝑛) = 𝑟𝑛 + 1 for all 𝑛 ∈ ℤ.
Show that 𝜙1 is a homomorphism but that 𝜙2 is not.



𝜙1 (𝑚 + 𝑛) = 𝑟(𝑚 + 𝑛) = 𝑟𝑚 + 𝑟𝑛 = 𝜙1 (𝑚) + 𝜙1 (𝑛)
so 𝜙1 is a homomorphism.


𝜙2 (𝑚 + 𝑛) = 𝑟(𝑚 + 𝑛) + 1 = 𝑟𝑚 + 𝑟𝑛 + 1
𝜙2 (𝑚) + 𝜙2 (𝑛) = 𝑟𝑚 + 1 + 𝑟𝑛 + 1 = 𝑟𝑚 + 𝑟𝑛 + 2.
Thus 𝜙2 (𝑚 + 𝑛) ≠ 𝜙2 (𝑚) + 𝜙2 (𝑛).

So 𝜙2 is not a homomorphism.