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How many elements are in the kernel of the homomorphism $f:(\mathbb{Z}/154\mathbb{Z})^* \to (\mathbb{Z}/154\mathbb{Z})^* $ where $f(x)=x^5$? The group operation in this case is multiplication with identity element $1 \mod 154$, so the group we are considering is $((\mathbb{Z}/154\mathbb{Z})^*,\cdot,\overline{1})$. The kernel of a homomorphism is given by: $$\operatorname{Ker}(f):=\{ g\in G|f(g) =e_G\}$$ There is a theorem that states that for any element in a group, the order of that element is either infinite or it must divide the number of elements in the group. In $(\mathbb{Z}/154 \mathbb{Z})^*$ there are 154 elements and $5 \nmid 154$. Therefore there are no element of order 5 and because 5 is prime no element other than the identity satisfies $x^5=e$. Therefore the kernel of $f$ contains only the identity element $1 \mod 154$. Is this correct or are there more elements in the kernel? The reason for this question is that I am not sure if this argument alone is enough. Thanks in advance!

EDIT I realise that I completely forgot to add the $^*$ to the group $\mathbb(Z)/154\mathbb{Z}$ when I first wrote this question and subsequently in trying to determine the number of elements made a mistake. I realise now that I need to evaluate $\phi(154)$ to find the number of elements and see if $5|\phi(154)$ where $\phi$ is the Euler-Totient function. I am sorry for the confusion.

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  • $\begingroup$ Are you writing the operation in $\mathbb{Z}/154\mathbb{Z}$ as multiplication? $\endgroup$
    – Tobias Kildetoft
    Commented May 6, 2013 at 14:40
  • $\begingroup$ Yes thanks I added it in the question. $\endgroup$
    – Slugger
    Commented May 6, 2013 at 14:41
  • $\begingroup$ But this is not a group under the usual multiplication. You can write the operation multiplicatively, but it is not the usual multiplication. $\endgroup$
    – Tobias Kildetoft
    Commented May 6, 2013 at 14:43
  • $\begingroup$ I am not sure what question you are asking exactly. What I mean is that if we have two element in $a,b \in\mathbb{Z}/154\mathbb{Z}$ the operation defined on them is $a\cdot b = ab \mod 154$. So I guess we are multiplying residue classes under the laws of modular arithmetic. $\endgroup$
    – Slugger
    Commented May 6, 2013 at 14:45
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    $\begingroup$ quite often you'll find that groups are even written as $G$, where the binary operator and identity are assumed to be obvious. not particularly in favour of this, but it's something that happens nonetheless. either way, my comment was intended as more of a heads-up than an actual correction! $\endgroup$
    – Tim
    Commented May 6, 2013 at 15:15

2 Answers 2

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The problem is that $\mathbb{Z}_{154}^\star $ does not, as you've said, have $154$ elements. Note that $\varphi(154)=60$.

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  • $\begingroup$ Thanks for your input I now realize my mistake. However the conclusion was easy with my mistake as 5 and 154 are coprime. Now that this is not the case because $5|60$ I am unsure how to continue... $\endgroup$
    – Slugger
    Commented May 6, 2013 at 14:58
  • $\begingroup$ Is your Avatar, Confucius, Alexander? :) and +1 $\endgroup$
    – Mikasa
    Commented May 6, 2013 at 15:00
  • $\begingroup$ @BabakS. It's Yu the Great. :) $\endgroup$
    – Alexander Gruber ♦
    Commented May 6, 2013 at 15:01
  • $\begingroup$ @TeunVerstraaten Hint: the number is the same for all abelian groups with order $60$. $\endgroup$
    – Alexander Gruber ♦
    Commented May 6, 2013 at 15:03
  • $\begingroup$ @AlexanderGruber Hmm I still can't figure it out. for sure there are at least 5 elements in the Kernel, namely $x,x^2,x^3,x^4,x^5=e$. Now suppose that $y \in Ker(f)$. Then $y^5=x^5=e$. Can I use this to deduce $y=x$ or is that conclusion not right? $\endgroup$
    – Slugger
    Commented May 6, 2013 at 16:47
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Hint: It's probably easier to consider $$\mathbb Z_{154}^* \cong \mathbb Z_2^*\times\mathbb Z_7^*\times \mathbb Z_{11}^*$$

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  • $\begingroup$ Hey thanks for the idea, however I am not sure how I would proceed from here to get a final answer. $\endgroup$
    – Slugger
    Commented May 6, 2013 at 14:57
  • $\begingroup$ It means you can restrict the problem to counting the kernel of $x^5$ on $\mathbb Z_p^*$ for $p=2,7,11$, and then multiple. Turns out $x^5$ is $1-1$ when $p=2,7$, and we know which elements of $\mathbb Z_{11}^*$ have $x^5=1$. $\endgroup$
    – Thomas Andrews
    Commented May 6, 2013 at 15:00
  • $\begingroup$ Ok thanks, the problem is that I would first have to show $\mathbb Z_{154}^* \cong \mathbb Z_2^*\times\mathbb Z_7^*\times \mathbb Z_{11}^*$ first $\endgroup$
    – Slugger
    Commented May 6, 2013 at 15:56
  • $\begingroup$ True, if you don't know the general theorem which allows you to deduce this, then this particular answer might not be useful. $\endgroup$
    – Thomas Andrews
    Commented May 6, 2013 at 16:00
  • $\begingroup$ @Teun, the way to use this (+1) idea is to first observe that for example $4^5\equiv1\pmod{11}$. So if you can find an integer $x$ that has remainder $4$ when divided by $11$, remainder $1$ when divided by $7$ (so that we would also have $x^5\equiv1\pmod7$) and also odd (so that $x^5\equiv1\pmod2$), then you would be in business. Trial and error is good enough here. $\endgroup$
    – Jyrki Lahtonen
    Commented May 6, 2013 at 17:19

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