# $a \in G$. $b \in G$. $\operatorname{ord}(a) = 12$. If $a = b^3$. Prove $\operatorname{ord}(b) = 36$

This is Pinter 10.H.1.

Let $$a$$ denote an element of a group $$G$$.

Let $$a$$ have order $$12$$. Prove that if $$a$$ has a cube root, say $$a = b^3$$ for some $$b \in G$$, then $$b$$ has order $$36$$.

HINT: Show that $$b^{36} = e$$. Then show that for each factor $$k$$ of 36, $$b^k = e$$ is impossible. [Example: If $$b^{12} = e$$, then $$b^{12} = (b^3)^4 = a^4 = e$$.] Derive your conclusion from these facts.

OK, let's begin.

We are given:

$$\operatorname{ord}(a) = 12 \tag{1}$$ $$a = b^3 \tag{2}$$

Here are the factors of $$36$$:

$$2, 3, 4, 6, 9, 12, 18$$

As suggested by the hint above, for each $$k$$ of these, we need to show that $$b^k = e$$.

Let's start with $$k = 2$$. Assume:

$$b^2 = e \tag{3}$$

$$a = b^3$$

$$a = b^2 b$$

Substitute (3):

$$a = e b$$

$$a = b \tag{4}$$

OK, back to (3):

$$b^2 = e$$

Substitute (4):

$$a^2 = e$$

So we've disproved the case where $$k = 2$$. We continue in a similar way for the rest of the factors of $$36$$ to complete the exercise.

# Second approach

Alright. Let's take a different approach for $$k = 2$$.

$$a = b^3$$

This time, we'll raise both sides to the power of $$12$$:

$$a^{12} = (b^3)^{12}$$

$$a^{12} = b^{36}$$

$$a^{12} = (b^2)^{18}$$

Substitute (3):

$$a^{12} = (e)^{18}$$

$$a^{12} = e$$

This conclusion is compatible with (1).

It seems like the second approach which does not contradict (1) should not be possible. What am I missing? Is there a step I'm taking which is invalid?

Exercises 10.H.2 and 10.H.3 are similar to this one and I can use a similar approach to the alternative one above. So I think by understanding this one, it'll clear up issues in those exercises as well.

In 10.H.1, we are told that $$\operatorname{ord}(b) = 36$$. So we know that for each factor $$k$$ of 36, we need to show that $$b^k = e$$ is not true. However in 10.H.2, and 10.H.3, we are not given the order of $$b$$. So we need to show definitively, for each possible $$k$$, whether $$b^k = e$$ is possible or not. So the second approach above is problematic because it doesn't help us rule out certain values of $$k$$. I'm wondering, what specific aspect of this approach should be avoided in performing these exercises? My guess is, the misstep is at saying "raise both sides to the power of 12".

• If $a^2=e$, then $a^{12}=e$ as well. Commented Aug 17, 2019 at 7:52
• Why should it not possible to compatible with $(1)$? This only means this method does not work.
– xbh
Commented Aug 17, 2019 at 7:53
• Just because your specific line of reasoning in (2) didn't reach a contradiction, that doesn't mean that there isn't a contradiction hiding behind some other line of reasoning from the same assumptions. Commented Aug 17, 2019 at 7:55
• This deserves more upvotes for its detailed context and attempts. Commented Sep 8, 2019 at 21:07

There are two factors of $$36$$ missing: $$1$$ and $$36$$.

If $$\operatorname{ord}b=k$$, then $$b^k=e$$ and therefore $$(b^k)^3=e$$. But $$(b^k)^3=(b^3)^k=a^k$$. So, $$a^k=e$$ and, since $$\operatorname{ord}a=12$$, this implies that $$12\mid k$$.

Clearly $$b^{36}=e$$. Do we have $$b^{12}=e$$? No, because $$b^{12}=a^4\neq e$$. By the same argument, $$b^{24}\neq e$$. Therefore, $$\operatorname{ord}b=36$$.

If $$a^2=e$$ then $$a^{12} = e$$ also holds. But $$Ord(a) = 12$$, so the lowest power that results in $$e$$ is $$12$$, so $$a^2 \neq e$$ and $$a^{12} = e$$. So, the second thing is not wrong, but it doesn't proof anything as you haven't shown whether the lower powers equal e or not.