1. ## Normal Subgroup

Prove that A4(+)Z3 has no subgroup of order 18.

I know A4 does not have a subgroup of order 6 because the converse of Lagrange thereom is false. So my initial notion is that because 6 is a multiple of 18, it cannot exist. Any thoughts on how to make this a proof?

Thanks All.

2. Originally Posted by RoboMyster5
Prove that A4(+)Z3 has no subgroup of order 18.

I know A4 does not have a subgroup of order 6 because the converse of Lagrange thereom is false. So my initial notion is that because 6 is a multiple of 18, it cannot exist. Any thoughts on how to make this a proof?

Thanks All.
Yeah, that's the idea. You can use the fact that if $G \cong A_4 \times C_3$ and $H \leq G$ then $H \cong H_1 \times H_2$ where $H_1 \leq A_4$ and $H_2 \leq C_3$. Essentially, that every subgroup of $A_3 \times C_3$ is a direct product of subgroups from these respective groups.

However, you should probably look at your proof of $A_4$ not having a subgroup of order 6 as $|A_4| = 12$, and 6 divides 12 (of course, I may be getting the wrong end of your stick here)...(if I was proving it, I'd look at what the elements of [tex]A_4[\math] look like).

3. Whats C3? Thats the chapter the problem is in so I thought maybe somehow normal sbgroups was suppose to be involved.

4. Originally Posted by Swlabr
Yeah, that's the idea. You can use the fact that if $G \cong A_4 \times C_3$ and $H \leq G$ then $H \cong H_1 \times H_2$ where $H_1 \leq A_4$ and $H_2 \leq C_3$. Essentially, that every subgroup of $A_3 \times C_3$ is a direct product of subgroups from these respective groups.

Why is this true in this case? This is definitively not so in other cases, for example: in the Klein group $\mathbb{Z}_2 + \mathbb{Z}_2$, the sbgp. $<(1,1)>=\{(0,0),(1,1)\}$ is not the direct sum of sbgps. of $\mathbb{Z}_2$

Tonio

However, you should probably look at your proof of $A_4$ not having a subgroup of order 6 as $|A_4| = 12$, and 6 divides 12 (of course, I may be getting the wrong end of your stick here)...(if I was proving it, I'd look at what the elements of $A_4$ look like).

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5. Originally Posted by tonio
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6. Originally Posted by RoboMyster5
Prove that A4(+)Z3 has no subgroup of order 18.

I know A4 does not have a subgroup of order 6 because the converse of Lagrange thereom is false.

Well, that is not a reason but in fact a consequence: $A_4$ has no sbgp. of order 6 because it has none (Add proof here)...

Here's an idea to prove what you want: suppose $H\,,\,|H|=18$ is a sbgp. of $G=A_4\times C_3$.

Since $|HA_4|=\frac{|H|\cdot |A_4|}{|H\cap A_4|}\le 36$, we get that $|H\cap A_4|\ge \frac{18\cdot 12}{36}=6\Longrightarrow$ The only option is $|H\cap A_4|=6$ (why??) , which contradicts that $A_4$ has no sbgp. of order 6 .

Tonio

So my initial notion is that because 6 is a multiple of 18, it cannot exist. Any thoughts on how to make this a proof?

Thanks All.
.

7. I'm still not comprehending what C3 is, but I am following your idea with the orders.

8. Originally Posted by RoboMyster5
I'm still not comprehending what C3 is, but I am following your idea with the orders.

$C_3$ is the cyclic group of order 3. It is the same (or better, isomorphic to) as $\mathbb{Z}_3$, but the notation $C_3$ is used when we want to remark the group is a multiplicative one, as opposed to the additive $\mathbb{Z}_3$

Tonio

9. Oh okay now this makes a lot more sense, wow thank you so much.