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Consider the subgroup $\mathbb{U}_{100}$ of the multiplicative group $\mathbb{C}^\times$ consisting of all the 100-th roots of unity. Define a group homomorphism as: $$ f:\mathbb{U}_{100}\to\mathbb{U}_{100},\space\space\space\space z\mapsto z^{70} $$

What is the order of the image of $f$?

I know that the image of $f$ is a set of the elements $\mathbb{U}_{100}$ such that the set is an image under $f$ but I'm not quite sure what to do...

Shaun
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  • Did you try to write an expression of an arbitrary element of $\mathbb U_{100}$ and plug that into the formula? If so, what happened? – Lee Mosher Nov 25 '20 at 17:24
  • Have you tried seing what happens if you replace 70 with 7? How about replacing 70 with 10? – Dan Rust Nov 25 '20 at 17:29

4 Answers4

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Hint:

The order of the image of a group homomorphism

is the order of the group divided by the order of the kernel.

J. W. Tanner
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The elements of $\mathbb U_{100}$ are $u^k$ where $u$ is the first $100$th root of unity. $$(u^k)^{70}=(u^{10})^{7k}$$ and $u^{10}$ is the first tenth root of unity. Powering this by any exponent will only give other tenth roots of unity, so the image of $f$ has order $10$. (Note that $7$ is coprime to $10$.)

Parcly Taxel
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  • Thanks! So if, for example, we had $z^{15}$ would it be $(u^k)^{15}=(u^{15})^k$ hence the order is $15$? –  Nov 25 '20 at 17:45
  • @butterchi $z^{15}$ would give an order of $20$, since $(u^k)^{15}=(u^5)^{3k}$. – Parcly Taxel Nov 25 '20 at 17:48
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Hint:

We see that $$ f(\mathbb{U}_{100})=\{ z^{70} | z \in \mathbb{U}_{100} \}=\{ e^{ik\pi\frac{7}{5}} | k \in [0,99]\}=\left < e^{i\pi\frac{7}{5}}\right>. $$ We see that $o \mid 10$ where $o$ is the order of $f(\mathbb{U}_{100})$. So $o \in \{1,2,5,10\}$. You can try each case and see that $o=10$.

Michelle
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In any group we have $\langle g^k \rangle = \langle g^d \rangle$ for $d=\gcd(k,n)$. One inclusion is obvious. The other follows from Bézout's identity.

Let $g$ be a primitive $100$th root of unity. Then $\mathbb{U}_{100} = \langle g \rangle$ and so the image of $f$ is $\langle g^{70} \rangle = \langle g^{10} \rangle$ since $10=\gcd(70,100)$. The order of $g^{10}$ is $100/10=10$ and so the image has order $10$.

lhf
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