Let $\varphi: G\rightarrow G'$ be a homomorphism. Show that for each $a \in G$, $| \varphi (a) | \le | a |$.

Question

Let $(G,*)$ and $(G',\circ)$ be groups. The order of an element $x$ of a group is denoted by $|x|$. Let $\varphi: G\rightarrow G'$ be a homomorphism. Show that, for each $a \in G$, $$\big| \varphi (a) \big| \le | a |\,.$$

I thought so, if $\varphi$ is a homomorphism then given $a, b \in G$, we have \begin{equation} \varphi (a \ast b) = \varphi (a) \diamond \varphi (b). \end{equation}

I don't know how to use this to show that the order of an element in the image is less than or equal to the order of that element. Can someone help me please?

Answer 1

Hint: We have $a^{|a|} = 1$, hence $1=\varphi(1) = \varphi(a^{|a|}) = \varphi(a)^{|a|}$. What can you conclude about $|\varphi(a)|?$

Answer 2

By induction, show that $\varphi(a^n) = \varphi(a)^n$ for all $n \in \mathbb Z$. Hence $$1_{G'} = \varphi(1_G) = \varphi(a^{|a|}) = \varphi(a)^{|a|}$$ and then...

Answer 3

Hint:

You have a surjective morphism from $\langle\mkern 2mu a\mkern2mu\rangle$ onto $\langle\, \varphi(a)\,\rangle$.