Abstract Nonsense

Crushing one theorem at a time

Review of Group Theory: Homomorphisms

Point of post: In this post we review basic facts about homomorphisms. Simple stuff like the image of a subgroup is a subgroup, the inverse image of a subgroup is a subgroup, etc. etc.


Just as linear transformations are the structure preserving maps for vector spaces homomorphisms are the structure preserving maps for groups. They, in essence, preserve most the qualities that interest us in the theory of groups. And, if the homomorphism happens to be bijective then we get a group isomorphisms (not to be confused with a linear isomorphism) which preserve ‘all’ the qualities that interest us. So, let us begin


Let \left(G,\star\right) and \left(G',\ast\right) be groups. Then a map \theta:G\to G' is a group homomorphism if

\theta\left(g_1\star g_2\right)=\theta\left(g_1\right)\ast\theta\left(g_2\right)

for all g_1,g_2\in G. We denote the set of all homomorphisms G\to G' by \text{Hom}\left(G,G'\right). We prove the immediate theorems

Theorem: Let G,G', and G'' be groups and \theta:G\to G' and \phi:G'\to G'' be homomorphisms then:

\displaystyle \begin{aligned}&\mathbf{(1)}\quad \theta(e_G)=e_{G'}\\ &\mathbf{(2)}\quad \theta\left(g^z\right)=\theta(g)^z,\text{ }z\in\mathbb{Z}\\ &\mathbf{(3)}\quad H\leqslant G\implies \theta\left(H\right)\leqslant G'\\ &\mathbf{(4)}\quad H\leqslant G'\implies \theta^{-1}\left(H\right)\leqslant G\\ &\mathbf{(5)}\quad\textit{If }\theta\textit{ is bijective, then }\theta^{-1}\textit{ is a homomorphism}\\ &\mathbf{(6)}\quad \textit{If }G=\left\langle S\right\rangle\textit{ and }\theta,\phi:G\to G'\textit{ are homomorphisms such that }\theta_{\mid S}=\phi_{\mid S}\textit{ then }\theta=\phi\\ &\mathbf{(7)}\quad\textit{If }G\textit{ is abelian and }\theta\textit{ surjective, then }G'\textit{ is abelian}\\ &\mathbf{(8)}\quad\textit{If }G\textit{ is cyclic, then }\theta(G)\textit{ is cyclic}\\ &\mathbf{(9)}\quad\phi\circ\theta\textit{ is a homomorphism}\end{aligned}


\mathbf{(1)}: This is immediately clear since \theta(e_G)=\theta(e_Ge_G)=\theta(e_G)\theta(e_G) and so upon cancellation \theta(e_G)=e_{G'}.


\mathbf{(2)}: It’s clear by induction that \theta\left(g^n\right)=\theta(g)^n for all n\in\mathbb{N}. Note though that by \mathbf{(1)} we have that


and thus \theta\left(g^{-1}\right)=\theta(g)^{-1}. Thus, it follows that


for all n\in\mathbb{N} from where the problem easily follows.


\mathbf{(3)}: Let \theta(g),\theta(h)\in \theta\left(H\right) then


Thus, noting that \theta(e_G)=\theta(e_{G'})\in \theta(H) we may conclude from a previous theorem that \theta(H)\leqslant G'.


\mathbf{(4)}: Suppose that g,h\in\theta^{-1}\left(H\right), then

\theta\left(gh^{-1}\right)=\theta(g)\theta(h)^{-1}\in H

so that gh^{-1}\in\theta^{-1}\left(H\right). Noticing then that e_G\in\theta^{-1}(\{e_{G'}\})\subseteq\theta^{-1}\left(H\right) allows us to conclude.


\mathbf{(5)}: This is similar to the proof that set-theoretic inverses of linear transformations are linear transformations. Namely, let g',h'\in G'. Then, since \theta is surjective there exists g,h\in G such that \theta(g)=g' and \theta(h)=h'. Then,


and thus


from where the conclusion follows.


\mathbf{(6)}: This follows since if g\in G is arbitrary there exists s_1,\cdots,s_n\in S such that

\displaystyle s_1^{\epsilon_1}\cdots s_n^{\epsilon_n}=g

where \epsilon_n\in\{-1,1\}. Then,

\displaystyle \begin{aligned}\theta(g) &= \theta\left(s_1^{\epsilon_1}\cdots s_n^{\epsilon_n}\right)\\ &= \theta(s_1)^{\epsilon_1}\cdots\theta(s_n)^{\epsilon_n}\\ &= \phi(s_1)^{\epsilon_1}\cdots\phi(s_n)^{\epsilon_n}\\ &= \phi\left(s_1^{\epsilon_1}\cdots s_n^{\epsilon_1}\right)\\ &= \phi(g)\end{aligned}

from where the conclusion follows.


\mathbf{(7)}:  Let g',h'\in G' be arbitrary. Since \theta is surjective there exists g,h\in G such that \theta(g)=g' and \theta(h)=h'. So,

\displaystyle g'h'=\theta(g)\theta(h)=\theta(gh)=\theta(hg)=\theta(h)\theta(g)=h'g'

since g',h' were arbitrary the conclusion follows.


\mathbf{(8)}: We merely note that

\displaystyle \begin{aligned}\theta\left(\left\langle g\right\rangle\right) &= \left\{\theta\left(g^z\right):z\in\mathbb{Z}\right\}\\ &= \left\{\theta(g)^z:z\in\mathbb{Z}\right\}\\ &= \left\langle \theta(g)\right\rangle\end{aligned}


\mathbf{(9)}: This follows from noticing that for any g,h\in G we have that





1.  Lang, Serge. Undergraduate Algebra. 3rd. ed. Springer, 2010. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print


December 28, 2010 - Posted by | Algebra, Group Theory, Uncategorized | , ,


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