# Homomorphism of Groups and divisibility of orders

I've been studying for my exams and bumped across this one particular question that I've been having a tad difficulty on:

Suppose that $\phi$ is a homomorphism from a finite group $G$ onto $H$ and that $H$ has an element of order $x$. Prove that $G$ has an element of order $x$ as well.

I tried emailing my prof and I was told to go about doing it first on a particular case when $x = 8$.

I made use of the fact that $\phi$ is onto so $\forall y \in H, \exists x \in G$ such that $\phi (x) = y$. So pick a particular $y$ such that $|y| = 8$ so we have $\phi(x)^8 = \phi(x^8) = y^8 = e$. Then by property of homomorphism we have that the order of $x$ is divisible by 8. Furthermore we have $x^8 \in \ker \phi$. This was about as far as I got and was hoping I can get help to finish this and thus the general case as well.

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There's a mistake in the last paragraph; you say that "the order of $x^8$ is divisible by 8", but this doesn't follow from what you've written. – Zev Chonoles Apr 10 '11 at 1:22
@Zev Corrected. I believe it should've said order of x is divisible by 8 since phi(x) = y. – Room Apr 10 '11 at 1:32
+1 for showing your work. – Arturo Magidin Apr 10 '11 at 3:30

You are indeed almost there. You know that $x^8\in\mathrm{ker}(\phi)$, so $\mathrm{ker}(\phi)\cap\langle x\rangle$ is nontrivial, and a subgroup of $\langle x\rangle$. In fact, because you are assuming that the order of $y$ is exactly $8$, then no smaller power of $x$ may lie in $\mathrm{ker}(\phi)$. So that means that $\mathrm{ker}(\phi)\cap \langle x\rangle = \langle x^8\rangle$.

Now, consider the subgroup $\langle x\rangle$. This is cyclic of order $n$ for some $n$. Since $\langle x^8\rangle$ is a subgroup, its index in $\langle x\rangle$ is a divisor of $n$. What's more: $8$ is the smallest power of $x$ that lies in $\langle x^8\rangle$; while this is not true for arbitrary cyclic groups (for example, if the order of $x$ were $12$, then $x^4 = x^{12}x^4 = x^{16}=(x^8)^2$ would also be in $\langle x^8\rangle$), here we know that the smallest power of $x$ that lies in $\mathrm{ker}(\phi)$ is $x^8$, and that $\langle x^8\rangle$ equals the intersection of $\langle x\rangle$ with $\mathrm{ker}(\phi)$. That means that the index of $\langle x^8\rangle$ in $\langle x\rangle$ is exactly $8$, and therefore that the order of $\langle x\rangle$ is a multiple of $8$. Since $n$ is the order of $x$, then the order of $x$ is a multiple of $8$, say $n=8k$. Then what is the order of $x^k$?

Now repeat the argument for the general case.

Added. In fact, you don't need to go that far down. You had already figured out that the order of $x$ is a multiple of $8$ (a simple way to see this: if the order of $x$ is $n$, and $\phi(x)=y$ has order $8$, then $y^n = \phi(x)^n = \phi(x^n) = \phi(e) = e$, so $8$, the order of $y$, divides $n$). So write $n=8k$. Then just note that since the order of $x$ is $8k$, then the order of $x^k$ is $8$, the same as the order of $y$.

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Can you explain again the part about the index? I am not fond of that terminology used in group theory in terms of what I've learned in my course and text book, unless you mean index in terms of cosets. In that context, from what I'm understanding the index of <x^8> in <x> is exactly 8 is the number of distinct left cosets of <x^8> in <x>. Again given that, what can I do from there? How can we say the order of <x> is a multiple of 8 given that fact? I think I kind of see it, yet I cannot! – Room Apr 10 '11 at 4:27
@Room: Suppose $G$ is any finite group, and $H$ is a subgroup. Say $|G|=n$, and $|H|=k$, and let $\ell$ be the index of $H$ in $G$; that is, the number of distinct left cosets of $H$ in $G$. Since every coset of $H$ has $k$ elements, and any two cosets are disjoint, every element of $G$ is in one and only one coset. So the number of elements of $G$ equals the number of cosets times the number of elements in each coset: $n = k\ell$. So $\ell$, the index of $H$ in $G$, divides $n$, the order of $G$. (This is just the other observation that is made from the argument that proves Lagrange's Thm). – Arturo Magidin Apr 10 '11 at 4:36
To your answer, given that, we can easily see that $x^n = x^{8k} = (x^k)^8 = e$. So we also see that the order of $x^k$ divides 8. Overall we now have that k divides 8, and 8 divides n and k divides n. Maybe there is some transitivity here but I am at a bit of a loss. – Room Apr 10 '11 at 4:56
@Room: The order of $x^k$ must be exactly 8; because if the order of $x^k$ is $r$, then $x^{kr} = e$, so $n$ divides $kr$ (since the order of $x$ is $n$); but $n=8k$, so if $8k$ divides $kr$, then $8$ divides $r$. Thus, the order of $x^k$, $r$, is a multiple of $8$. You know it's also a divisor of $8$ (by the above), so it is equal to $8$. More generally, the order of $x^i$ is equal to $n/\gcd(n,i)$ for any integer $i$. Since $n=8k$, then $\gcd(n,k) =k$, so $n/\gcd(n,k) = n/k = 8$. – Arturo Magidin Apr 10 '11 at 4:59
Beautiful. Thank you very much! – Room Apr 10 '11 at 5:03

You're almost there!

Because now, consider the group $\langle x\rangle\leq G$ generated by $x$. This group is cyclic, what do you know about its order?

Finally, use that finite cyclic groups of order $n$ have at least (in fact precisely) one cyclic subgroup of order any divisor of $n$.

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How do we know that the the group generated by x is a sub"set" of ker(phi) ? I might be missing some property about homomorphism if so. – Room Apr 10 '11 at 1:35
Sorry, you don't know that indeed, I don't know why I wrote that. The point I'm trying to make is that $\langle x\rangle$ is in the finite group $G$. – Myself Apr 10 '11 at 1:38