# Proving that $\varphi$ is a homomorphism

Could someone please help me answer this question? Consider the function $\varphi:\mathbb{Z}_{30}^*\to\mathbb{Z}_{30}^*$ given by the formula $\varphi(x)=x^2$. Prove that $\varphi$ is a homomorphism and compute its kernel.

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You need enclose the $\LaTeX$ expressions in dollar signs, e.g., $\varphi(x)=x^2$. – Brian M. Scott Feb 2 '13 at 0:29
For any abelian group $\,A\,$ and for any integer $\,n\,\,,\,f_n:A\to A\,\,,\,f_n(x):=x^n\,$ is an endomorphism of $\,A\,$... – DonAntonio Feb 2 '13 at 0:32
@Swastika Devi I added the tag "Group Theory" to your question. You can use more than one tag when you pose a question and you should use at least one tag to place the question is some specific area of maths. – Ittay Weiss Feb 2 '13 at 0:45

It would help you if you first identify the elements in $\mathbb {Z}^*_{30}$, you will find there are not too many. Next, you should know the definition of homomorphism: A function $\varphi:G\to H$, where $G,H$ are groups, is a homomorphism if for all $g_1,g_2\in G$ holds that $\varphi(g_1g_2)=\varphi(g_1)\varphi(g_2)$. Now, in your case it is given that $G=H=\mathbb Z^*_{30}$ and that $\varphi(x)=x^2$ for all $x\in \mathbb Z^*_{30}$. So, you have a very explicit formula to work with and check that it indeed holds.

For instance, take a couple of values $x,y\in \mathbb Z^*_{30}$ and compute $\varphi(x)\varphi(y)$, and then compute $\varphi(xy)$. Do you get the same result? (You should!). Remember that the operation is multiplication modulo $30$. Now, since there are not too many elements to check you can just check all of them. But, that will be tiresome and inefficient. Can you think of a general argument to show that the defining equality of a homomorphism is satisfied in this case? (Hint: start your general proof by "Let $x,y\in \mathbb Z^*_{30}$" and then state what equality needs to be proven in terms of $\varphi$ and then use the formula for $\varphi$.)

Once this is done, to compute the kernel you need to know what the definition of the kernel is. I'm sure you can find it in the notes you use that $ker(\varphi)=\{x\in \mathbb Z^*_{30} \mid \varphi(x)=e\}$. What is $e$ in your case? What is $\varphi(x)$? Can you find all values $x\in \mathbb Z^*_{30}$ that satisfy $\varphi(x)=e$? That will be the kernel.

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That it is a homomorphism (using that the multiplicative group is abelien):

$$\varphi(xy) = (xy)^2 = xyxy = x^2y^2 = \varphi(x)\varphi(y).$$

Finding the kernel: You really just want to solve $x^2 = 1$. Try by hand to take the square of each element in $\mathbb{Z}_{30}^*$. The kernel is all those $x$ such that $x^2 = 1$ (all this is modulo $30$ remember). For example $7^2 = 49 = 19 \neq 1$, so $7$ is not in the kernel. So this of course assumes that you have found all the elements in $\mathbb{Z}_{30}^{*}$.

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warning: 10 is also not in $\mathbb Z^*_{30}$ so the last sentence is confusing. – Ittay Weiss Feb 2 '13 at 0:43
@IttayWeiss: Thanks for pointing out the mistake. – Thomas Feb 2 '13 at 1:40

This is a finite set, you could check all the relevant cases. By hand, even. Or set up the equations and check they are fullfilled. Dig in! Try some ideas, see where they lead you. This is homework, designed for you to learn by (trying to) do it.

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Thank You Sir.Let me go back to lecture notes again and check. – Swastika Devi Feb 2 '13 at 0:58