Suppose I am given $G_1,H_1$ as groups and $f_1: G_1 \to H_1 $ a group homomorphism. Then $$\ker f_1 := \{g_1 \in G_1 : f_1(g_1) = id_{H_1}\} \tag{1}$$

Suppose I am given $G_2, H_2$ as vector spaces and $f_2: G_2 \to H_2 $ a linear transformation. Then $$\ker f_2 := \{\mathbf{g_2} \in G_2 : F_2(\mathbf{g_2}) = 0_{H_2}\} \tag{2} $$

If $H_2$ is a vector space, then $id_{H_2}$ is the identity linear transformation. But $ 0_{H_2} $ is $ 0(\mathbf{h_2}) = \mathbf{0} \text{ } \forall \mathbf{h_2}\in H_2. $ so it's different! Shouldn't $0_{H_2}$ in (2) really be $id_{H_2}$ instead?

Edit 2 hours later: I'm working with Möbius maps and the textbook says

"The kernel of the homomorphism $\phi: GL(2, \mathbb{C}) \to \text{(Möbius group)} $ via $$ \phi: \left[ \begin{matrix} a & b \\ c & d \\ \end{matrix} \right] \to f(z) = \dfrac{az + b}{cz + d} \in \mathbb{C} $$ is the set of matrices $\mathbf{A}$ such that the Möbius map $\phi(\mathbf{A})$ is the identity map. So to find the kernel, solve for $z$ in $f(z) = \dfrac{az + b}{cz + d} = z $.

Thanks to all of you, I now understand that $id_{H_1}$ in (1) means the identity element in $H_1$, NOT identity map. But why does the textbook say "map" then? Shouldn't it say "element"? I understand the identity element in the Möbius group is $f(z) = z$.

  • $\begingroup$ $id_{H_1}$ as used here is not standard notation (and for a good reason, as shown by your question...). Usually it'd be either $1_{H_1}$ or $e_{H_1}$. $\endgroup$
    – tomasz
    Jan 1, 2013 at 22:20
  • 1
    $\begingroup$ "Why does the textbook say 'map' then?" Because in the Mobius group, the elements are maps (from $\mathbb{C}$ to $\mathbb{C}$). The identity element of the Mobius group is the identity map from $\mathbb{C}$ to $\mathbb{C}$. These "maps" are not maps from one group to another (like your $f_1$ and $f_2$ are), but simply the objects of a single group. $\endgroup$
    – Ted
    Jan 2, 2013 at 1:46

2 Answers 2


Your $id_{H_1}$ is not the identity map, but rather the identity element of the group $H_1$, often denoted $1$ instead of $id$. And for abelian groups, written additively, one usually denotes this group element as $0$. In case of linear maps, we deal with vector spaces, which are especially abelian groups (under vector addition), hence the use of $0$.

Indeed, any linear map between vector spaces is also a homomorphism of the underlying (abelian) groups.

I think you got confused between identity (or neutral) element of a group and the identity map (of a set or whatever). Also be sure not to mix the $0$ vector of a vector space with the $0$ of the field or the zero map that maps every element of some set to $0$, even though we commonly use the very same symbol for all of these, hoping that the context should make clear which of these is meant (beacause e.g. the other side of an equation demands a vector, a scalar, or a vector- or scalar-valued map).


You are mixing two different things: when you write $id_H$ you mean the identity element of $H$, i.e. the element of $H$ such that for all $h\in H$ we have $h*id_H=id_H*h=h$. Such an element is $0$ in a vector space.
$0_{H_2}$ does not represent the $0$ transformation but rather the zero element of a vector space.


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