I have a group $G:=\mathbb{R}^2 = \{(a,b)|a,b \in \mathbb{R}\}$ where $(a,b)+(c,d) =(a+c,b+d)$ and a subgroup $H: = \{(a,a)|a,a \in \mathbb{R}\}$.

I now have to find a group homomorphism $\phi : \mathbb R^2 \to \mathbb R$, such that $\ker \phi = H$. Where the group operation on $\mathbb{R}$ is the usual addition.

I know that I have to finde something that satisfies the following:

Let $(G_1, \cdot_1)$ and $(G_2, \cdot_2)$ be two groups. A function $\phi : G_1 \to G_2$ is called a group homomorphism, if it satisfies

  1. $\phi(e_1) = e_2$, with $e_1$ the identity element of $G_1$ and $e_2$ the identity element of $G_2$,

  2. $\phi(f \cdot_1 g) = \phi(f) \cdot_2 \phi(g)$.


$\ker \phi := \{g \in G_1\, |\,\phi(g) = e_2\}$.

  • $\begingroup$ You're considering the wrong operations: $\mathbf R$ and $\mathbf R^2$ are not groups for mulrtplication. $\endgroup$
    – Bernard
    Dec 14, 2015 at 1:02
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    $\begingroup$ The $\cdot_1$ does not stand for multiplikation but for the action used on the group in this case the usual addition. $\endgroup$
    – Sofie
    Dec 14, 2015 at 1:04
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    $\begingroup$ @Sofie: If the group operation is addition, then $e_2=0$, not $1$. (HINT: Think what axiom the identity element satisfies . . .) $\endgroup$ Dec 14, 2015 at 1:05
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    $\begingroup$ Because $\mathbb{R}$ is an additive group, it's usual to use $0$ as the symbol for its identity element @Sofie. I recommend changing it to help others understand your question more easily. $\endgroup$ Dec 14, 2015 at 1:06
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    $\begingroup$ Hint: look for homomorphisms of the form $\phi(a,b)=a*M+b*N$ for suitable real numbers $M,N$. (Check that those are all homomorphisms!). Now you just need to find $M,N$ such that this vanishes on $H$. $\endgroup$
    – lulu
    Dec 14, 2015 at 1:08

1 Answer 1


Consider $\phi:\mathbb{R}^2\to\mathbb{R}$ defined by $\phi(a,b)=a-b$. Then $\phi(0,0)=0-0=0$ and $$\phi((a,b)+(c,d))=\phi(a+c,b+d)=(a+c)-(b+d)=(a-b)+(c-d)=\phi(a,b)+\phi(c,d)$$

Furthermore, $ker(\phi)=\{(a,b): \phi(a,b)=0\}=\{(a,b): a-b=0\}=\{(a,b): a=b\}=H$

  • $\begingroup$ This is even simpler, thank you! $\endgroup$
    – Sofie
    Dec 14, 2015 at 1:32

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