2
$\begingroup$

I am trying to do an exercise but I am confused at this question. Isn't the order of $a$ just $3$.

I am trying to use Lagrange Theorem but to no avail since I have no clue what $G$ is.

Order Group Question

$\endgroup$
  • $\begingroup$ Well, at least $\;30\;$ , uh? $\endgroup$ – DonAntonio Dec 12 '17 at 0:31
  • $\begingroup$ @DonAntonia How did you get that ? I understand that 3 divides the order of the factor group and order of G divides 10 thus 30 is number of elements in G at least but how does that pertain to order of a ? $\endgroup$ – Nameless Dec 12 '17 at 0:32
  • $\begingroup$ So, offhand, you know that $a^3H = H$, not that the order of $a$ is $3$. $\endgroup$ – Ted Shifrin Dec 12 '17 at 0:33
  • 1
    $\begingroup$ @DonAntonio - what about $a =(1,0)\in {\mathbb Z/3}\times {\mathbb Z/10}$? $\endgroup$ – peter a g Dec 12 '17 at 0:34
  • 2
    $\begingroup$ Let $m$ be the order of $a$.Then $(aH)^{m} = a^{m}H = H$. It follows that $3|m$ so that $m = 3k$. $\endgroup$ – akech Dec 12 '17 at 1:19
3
$\begingroup$

$|a|$ clearly divides $30$, for we have:

$a^{30} = (a^3)^{10} = h^{10}$ for some $h \in H$, and by Lagrange, $h^{10} = e$ (since $\langle h\rangle$ is a subgroup of $H$ and its order, which is $|h|$, must divide $10$).

On, the other hand, if $|a| = k$, then $a^k = e \in H$, and since $aH$ has order $3$, we must have $3 \mid k$ (for otherwise, $aH = H$, or $a^2H = H$, contradiction).

The possibilities are thus:

$3,6,15,30$.

Letting $G = \Bbb Z_{30}$ and $H = \langle 3\rangle$ (which has order $10$), we see that $1+H$ and $2+H$ both have order $3$ in $G/H$, and:

$10 \in 1+H$ has order $3$ in $G$.

$5 \in 2+H$ has order $6$ in $G$.

$2 \in 2+H$ has order $15$ in $G$, and

$1 \in 1+H$ has order $30$ in $G$, which settles the question.

$\endgroup$
2
$\begingroup$

Consider a subgroup $K\subseteq G/H$ generated by $aH$. It follows that $K\simeq G'/H$ for some subgroup $H\subseteq G'\subseteq G$ such that $a\in G'$. Since $|G'/H|=3$ and $|H|=10$ then it follows (by Lagrange's theorem) that $|G'|=30$. And since $a\in G'$ then $|a|$ divides $30$.

Are all divisors of $30$ possible? If you consider $G'=\mathbb{Z}_3\times\mathbb{Z}_{10}$ and $H=0\times\mathbb{Z}_{10}$ then $3, 6, 15, 30$ are clearly possible values of $|a|$. Obviously $1$ is not possible.

What about $2, 5$? (thanks to @DavidWheeler) Of course if $|a|=2$ or $|a|=5$ then $K$ being the image of $\langle a\rangle$ would have to be of order $1,2$ or $5$. But $|K|=3$. Contradiction.

$\endgroup$
  • $\begingroup$ 2 is not possible, for if so, $aH$ could not have order 3. 5 is likewise not possible for it leads to $H = a^3H = a^5H \implies a^2 \in H$. $\endgroup$ – David Wheeler Dec 12 '17 at 12:06
1
$\begingroup$

A variation on David W's answer.

Write $\langle a\rangle$ for the cyclic subgroup of $G$ generated by $a$.

One has a surjective homomorphism $\pi$

$$ \left< a \right> \to \langle a\rangle H / H,$$ defined by $\pi\colon a^k\mapsto a^k H$. This is a homomorphism, as $H$ is normal.

The kernel $K$ of $\pi$ is $\langle a\rangle \cap H$. By hypothesis the image of $\pi$ has cardinality $3$. Therefore $\#\langle a \rangle = 3 \ \cdot \#K$. Now, the only possibilities for the cardinality of $K$ are $1,2,5, 10$, as $K$ is a subgroup of $H$. So the only possibilities for the order of $a$ are $3, 6, 15, 30$. By explicit construction (as in the comments, or in David's answer), we see all of these occur.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.