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I'm reviewing my Abstract Algebra and I'm stuck on something. My professor explained that if $\varphi:G\rightarrow H$ is a homomorphism, then $$\varphi(1_G)=1_H$$where $1_X$ is the identity in the group $X$. What I'm confused about is the following: we were asked to count the number of homomorphisms from $Z_a$ to $Z_b$, where $a,b$ are positive integers and $a<b$.

At the time it made sense but thinking about it now I'm confused. $1_{Z_a}$ generates $Z_a$ and $1_{Z_b}$ generates $Z_b$. I know the number of such homomorphisms is $\gcd(a,b)$. Why is this not $1$ homomorphism? From what my professor told me if $\varphi$ is a homomorphism from $Z_a$ to $Z_b$ then $\varphi(1_{Z_a})=1_{Z_b}$. But then as everything in $Z_a$ is determined by $1_{Z_a}$, isn't the entire image of $Z_a$ under $\varphi$ determined already? He talked about mapping $1_{Z_a} \mapsto x \in Z_b$. But how can $x$ be anything but $1_{Z_b}$ by what he told us that same day!? I feel like I'm missing something very obvious and fundamental about group homomorphisms here, what is it? Thanks!

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    $\begingroup$ I think you are confusing the element 1 with 0. The mapping $1_{Z_{A}} \mapsto x$ refers to the element $1$ not $0$ which is the identity of the group. $\endgroup$ – Complexification Jan 18 '14 at 2:37
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You confuse $1$ with $1$. Note that $1_G$ represents the identity of $G$.

In $\mathbb{Z}_n$ the number $1$ (or the class of 1) is NOT the identity of the group. The identity is $0$.

So, $\varphi(1_G)=1_H$ actually says:

$$\varphi(0)=0 \,.$$

This is exactly why I prefer to denote $1_G$ by $e$ or $e_G$....

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  • $\begingroup$ Thanks! This made things so clear! $\endgroup$ – TinyTim Jan 18 '14 at 2:41
  • $\begingroup$ My professor liked to do everything in terms of what things 'would look like' in $\mathbb{R}$ and it could get confusing. $\endgroup$ – TinyTim Jan 18 '14 at 2:44
  • $\begingroup$ How is it possible to confuse $1$ and $1$? $\endgroup$ – David Jun 25 at 11:31
  • $\begingroup$ @David $1_G$ or simply $1$ is often used to denote the identity of $G$. This is a standard notation if the group operation is denoted multiplicatively (which is the standard choice for non-Abelian groups, or for groups not known to be Abelian). If the book uses $1$ instead of $e$ to denote the identity of the group, then $1$ suddenly can have two different meanings in the group $(\mathbb{Z}, +)$: the number 1 and the identity element "1" (which is actually the number 0). Using $e$ instead of $1$ for multiplicative groups would avoid this confusion. $\endgroup$ – N. S. Jun 25 at 11:38
  • $\begingroup$ @N.S. But your answer writes both indistinctly! $\endgroup$ – David Jun 25 at 13:20
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The confusion seems to be about multiplicative vs. additive notation. $1_G$ denotes the identity element of the group $G$ (which may or may not be numerically equal to $1$).

In this case $1_{Z_a}$ denotes the identity element in the group $Z_a$, so we have $1_{Z_a}=0$. So, $1_{Z_a}$ does not generate $Z_a$ (unless $a=1$).

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  • $\begingroup$ Thanks! This helped clear things up! $\endgroup$ – TinyTim Jan 18 '14 at 2:46
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Note that your $1_{\mathbb{Z}_{a}}$ is actually $0$ as you're taking the additive group. The $1$ is not the additive identity. Therefore, $1\mapsto x$ may be a homomorphism for $x\neq1$.

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  • $\begingroup$ Thanks! I feel like an idiot, but this is why I'm reviewing! $\endgroup$ – TinyTim Jan 18 '14 at 2:46
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Note that the homomorphism could be zero. If the homomorphism is non zero, then as you remarked, everything is determined by where $1$ goes. $1$ can go to any of the generators of $\mathbb{Z_m}$. Your confusion stems from the fact that you are confusing notations.

For a group written in "multiplicative notation" $1_G$ means the identity. For abelian groups like $Z_n$ we use additive notation, and here the identity is $0$ and not $1$

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