Group homomorphisms

We want to study different maps between groups, so called homomorphisms, in this section. Homomorphisms (and especially isomorphisms) are very useful and important tools for understanding structures and properties of groups, for understanding how we can take one group structure and see whether we can move it to the other group, or for classifying groups up to isomorphisms. The last few words basically mean that two groups can be considered the same, although they may have different elements and different operations, while the whole structures are identical.

Sources:

  • Drapal part 1.4
  • Wiki post dealing with group homomorphisms

Theory:

We already mentioned group homomorphisms in the previous post on cyclic groups. Recall that if you have two groups G(*_G,{^{-1}}_G,e_G) and H(*_H,{^{-1}}_H,e_H), and a map \varphi:G\to H , then \varphi is a homomorphism, if \varphi(x{\cdot}_G y)=\varphi(x){\cdot}_H\varphi(y) for any two x,y\in G. If \varphi is also a bijection, then we say that \varphi is an isomorphism (denoted by \cong).

Properties:

  • If \varphi:G\to H is a group homomorphism, then \varphi(e_G)=e_H and \varphi(x^{-1})=\big(\varphi(x)\big)^{-1}.
  • \varphi:G\to H is an isomorphism if and only if \varphi is a homomorphism and there exists a homomorphism \psi:H\to G such that \varphi\circ\psi=id_H and \psi\circ\varphi=id_G.
  • Between any two groups G and H there always exists at least one homomorphism \varphi:G\to H, namely the zero homomorphism sending each element g\in G to e_H\in H.  
  • Let n=d\cdot r. Then d\cdot\mathbb{Z}_n\cong\mathbb{Z}_r.
  • For every a\in\mathbb{Z}_n, a map \varphi_a:\mathbb{Z}_n\to\mathbb{Z}_n; i\mapsto a\cdot i, is a group homomorphism.
  • Every group homomorphism \varphi:\mathbb{Z}_n\to\mathbb{Z}_n is equal to \varphi_a for some a\in\mathbb{Z}_n.
  • \varphi_a is an isomorphism (i.e., an automorphism) \iff (a,n)=1.
  • For every 0\neq a\in\mathbb{Z}_n we have a\cdot \mathbb{Z}_n=d\cdot\mathbb{Z}_n, where d=(a,n).

Examples:

Example 1:

Let G be a cyclic group generated by an element a, such that a^{10}=1 and a^i\neq  e_G, i=\{1,…,9\}. Show that G is isomorphic to H=\mathbb{Z}_{10}.

Solution:

Realise that G=\{a,a^2,a^3,\ldots,a^9,a^{10}=1\}. Consider a map \varphi:G\to H given by a^i\mapsto i. We will show that \varphi is an isomorphism, i.e., a bijective homomorphism.

  • Homomorphism: Take arbitrary two elements of G, namely a^I, a^j and consider their product a^i\cdot a^j=a^{i+j\ mod\ 10}. Then

        \[\varphi(a^i\cdot a^j)=\varphi(a^{i+j\ mod\ 10})=(i+j)\ mod\ 10\]

        \[=\big(\varphi(a^i)+\varphi(a^j)\big)\ mod\ 10=\varphi(a^i)+\varphi(a^j).\]

    By definition this proves that \varphi is a homomorphism.  
  • Bijection: \varphi is obviously surjective – given any i\in\mathbb{Z}_{10}, then \varphi^{-1}(i)=a^i\in G. Since \varphi is a surjective map between two finite sets of the same cardinality, it is also injective, and hence, a bijection.

Example 2:

Find all homomorphisms \varphi:\mathbb{Z}_5\to\mathbb{Z}_6.

Solution:

We know there is at least the zero homomorphism, but what about some other homomorphisms? Well, by the first property, \varphi(0)=0 and \varphi(1)=a for some a\in\mathbb{Z}_6. Also, we have

    \[0=\varphi(0)=\varphi(1+1+1+1+1)=5\cdot\varphi(1)=5\cdot a.\]

So, 5\cdot a=0. However, the equation 5\cdot a\equiv 0 mod 6 has only one solution, namely a=0. So, we see that there is no homomorphism other than the zero homomorphism.

Problems:  

Problem 1:

Prove the first of the properties of group homomorphism. (Hint: the identity element and inverses are unique in any group.)

Problem 2:

Let X be a set X=\{1,-1,i,-i\}. Show that X can be viewed as a multiplicative group G_X and find a group H\neq G, such that G\cong H.   

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