# Finding the non zero elements in $\mathbb{Z}_{30}^*$ and proving that a function is a homomorphism and computing its kernel

I have a question where I really need help and guidance :

1. please if someone can guide how can we find the non-zero elements of $\mathbb{Z}^*_{30}$

2. please guide how can I prove that the function $\varphi : \mathbb{Z}^*_{30} → \mathbb{Z}^*_{30}$ given by the formular $$\varphi(x) = x^2$$ is a homomorphism.

3. To compute its kernal, how can I use the idea that the kernal of the function is the inverse image of the trivial subgroup, that is e ≤ H

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Are you sure you are interested in $\mathbb Z_{30}$ and not $\mathbb Z_{30}^*$? Do you know the definition of homomorphism? – Ittay Weiss Jan 27 '13 at 23:28
1) What does finding "non-zero elements" in $Z_{30}$ mean? 2) Is it $\phi(x) = x^2$? – Isomorphism Jan 27 '13 at 23:29
please tag your question appropriately (add "group theory"). And, use latex coding for mathematics symbols. – Ittay Weiss Jan 27 '13 at 23:31
It seems somebody was kind enough to edit your question for you. You might want to hit the edit button to see the correct way to write the question and the latex code. – Ittay Weiss Jan 27 '13 at 23:36
${\bf Z}_{30}$ is a group under (modular) addition. $\phi(x)=x^2$ is not a homomorphism on this group. As suggested in the comments, maybe you want some other group related to ${\bf Z}_{30}$. – Gerry Myerson Jan 28 '13 at 0:27

There is a multiplicative group associated to $\mathbb{Z}_{30}$.

$2,3,5$ are all zero-divisors, so you need to rule them out. And their multiples.

$(\mathbb{Z}_{30})^* = 1,7,11,13,17,19,23,29$

Here,

• $7^2=23^2=49\equiv 19$
• $11^2=19^2=121\equiv1$
• $13^2=17^2=169\equiv 19$
• $29^2=1^2=1$

So the square map takes $\{ 1,7,11,13,17,19,23,29 \} \mapsto \{ 1, 19\}$.

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Thank you Sir for your guidance and help...but how can we prove that the function above given by the formula is a homomorphism..? – Denish Sen Feb 2 '13 at 0:29
Also, please guide and help, how can we compute its kernal..? – Denish Sen Feb 2 '13 at 0:35
The spelling is "kernel". The definition is, the elements that are mapped to the identity. Do you know which is the identity in this group? Can you apply the square map to the eight elements of the group to see which ones wind up at the identity? For proving it's a homomorphism, you have to use the definition of homomorphism, then you should be able to write down a proof that in any abelian group the squaring map is a homomorphism. Try it! – Gerry Myerson Feb 2 '13 at 5:27
Thank you Sir...for the help and guidance – Denish Sen Feb 2 '13 at 9:13