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Let $\phi : G \to H$ be a homomorphism. Show the following: if $H$ is abelian, then for every $x,y \in G$ the image of the element $xyx^{-1}y^{-1}$ is $1_H$. Write down the properties you use.


Here is what I did:

Choose $x,y \in G$ randomly. By the definition of groups (closed under multiplication and inverses) $xyx^{-1}y^{-1} \in G$.

$\phi(xyx^{-1}y^{-1})=\phi(x)\phi(y)\phi(x^{-1})\phi(y^{-1}) \in H$ by the definition of homomorphisms.

Since $H$ is abelian, we can write: $\phi(x)\phi(y)\phi(x^{-1})\phi(y^{-1})=\phi(x)\phi(x^{-1})\phi(y)\phi(y^{-1})$

Then, again by definition of homomorphisms: $\phi(x)\phi(x^{-1})\phi(y)\phi(y^{-1})=\phi(xx^{-1}yy^{-1})=\phi(1_G 1_G)=\phi(1_G)=\phi(1_G)\phi(1_G)$.

From the last equation it follows that $\phi(1_G)=1_H$.

So in conclusion we have:

$\phi(xyx^{-1}y^{-1})=1_H$

I would like for someone to check whether this is correct. Thanks!

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  • $\begingroup$ It is correct :). $\endgroup$ – Jennifer Jun 6 '16 at 14:16
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this is correct but a bit long, we say that you not know a property of homomorphism $f$, that is $f$ transform $y^{- 1}$ to $(f (y))^{- 1}$, a remark that the hypothesis $H$ abelian is not necessary, enough that the image of $f$ is.

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Your solution is correct. Another method is to use the fact that a homomorphism between groups maps inverses to inverses. In other words, if $\phi: G \rightarrow H$ is a homomorphism, then $\phi$ takes $x^{-1}$ to $\phi(x)^{-1}$. So, $\phi(xyx^{-1}y^{-1}) = \phi(x)\phi(y)\phi(x^{-1}) \phi(y^{-1}) = \phi(x) \phi(y) \phi(x)^{-1} \phi(y)^{-1}$. Now use the fact that $H$ is abelian to show that this product is $1_H$.

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