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The questions asks me to find all groups up to isomorphism of the semi direct product $C_5 \rtimes C_4$

Now I've done the working out to get four groups (Note I've used $X$ as an element of $C_5$ and $Y$ as an element of $C_4$ so in all of these $X^5=1$, $Y^4=1$):

a). $YX=XY$ - which abelian so it is just the group $C_{20}$

b). $YXY^{-1}=X^2$

c). $YXY^{-1}=X^4$

d). $YXY^{-1}=X^3$

Now I have found that by changing the generator $Y^3$ to say $W$ in (b) you get the group (d) so (b) and (d) are isomorphic. I also thought that the same applied to (b) and (c) when you change the generator in (b) from $Y^2$ to say $Z$ but in the answers this isn't the case and it only says that (b) and (d) are isomorphic. Can anyone tell me how to spot which groups are isomorphic and which aren't? My notes aren't very clear and I've tried to search online to no avail. Thanks.

Lolwat
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  • The possible non-tivial subgroups here have orders $,2, 4, 5,10,$ . Of these, only with $,10,$ we might have more than one group up to isomorphism (why?), so focus on this order . – DonAntonio Apr 29 '13 at 19:04
  • Thanks, but I don't understand why the non-trivial groups have order $2,4,5,10$. Can you clarify this for me please? – Lolwat Apr 29 '13 at 19:06
  • I didn't say that. I said that there is only one group up to isomorphism of orders $,2,5,$ , and all the groups of order $,4,$ must be isomorphic as they're Sylow $,2$-subgroups, so only with subgroups of order $,10,$ you could have more than one isomorphism class... – DonAntonio Apr 29 '13 at 19:09
  • Sorry I'm now completely lost here. I'm confused. How do you know what order the groups have? I'm struggling to find out how. – Lolwat Apr 29 '13 at 19:12
  • Can anyone help please? – Lolwat Apr 29 '13 at 19:41
  • I think DonAntonio was trying to suggest a way to tell that (b) and (c) are different. I would suggest looking at the centralizer of $Y^2$. Calculate $YYXY^{-1}Y^{-1}$ in (b) versus (c); it is a power of $X$ in both, but a different power. – Jack Schmidt Apr 29 '13 at 20:56
  • Not precisely, @JackSchmidt . The OP wrote he wants all the subgroups of the given semidirect product up to isomorphism. I stressed the fact that there are not many problems with the possible orders $,2,4,5,$ (all of which, btw, exist), since $,2,5,$ are primes and Sylow $;2$-subgroups are isomorphic, so finding one of each of these is finding all of them (again, up to isomorphism). With $,10,$ though there may be problems as there are two groups of this order up to isomorphism nad perhaps they both could appear as subgroups here, so I wrote him to focus on this case. – DonAntonio Apr 29 '13 at 22:05
  • @DonAntonio: I don't think the OP mentioned subgroups at all? – Jack Schmidt Apr 30 '13 at 02:12
  • @JackSchmidt, I think you're right. I must have misread, though funny enough my comments still can help him about this...Thanks. – DonAntonio Apr 30 '13 at 02:15

2 Answers2

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You can't change $y^2$ to a generator since $y^2$ isn't a generator of $\mathbb{Z}_4$. The group c is the binary dihedral group (the double of the symmetries of a 5 gon). Note that in this group $y^2$ is central since $yxy^{-1}$ flips $x$ around to its inverse, so if you do that twice you're back where you started. Meanwhile the other two non abelian guys actually have only the identity in their centers! Note that you now need to do $yxy^{-1}$ not twice, but four times to get back where you started. Think of $y$ as playing the role of multiplication by $-1$ in $\mathbb{Z}_5$ in c.) while playing the role of multiplication by $2$ in $\mathbb{Z}_5$ in the other two (b and d). If you multiply by $-1$ twice you get $1$ (${(-1)^2=1 \bmod 5}$), but you need to do it four times since in $\mathbb{Z}_5$, ${2^4=16 = 1 \bmod 5}$, but ${2^2=4 \neq 1 \bmod 5}$.

In general, you can show that for $p$ prime, ${\mathbb{Z}_p \rtimes_a \mathbb{Z}_n}$ (the second generator acts as multiplication by $a$ on the first) and ${\mathbb{Z}_p \rtimes_b \mathbb{Z}_n}$ (the second generator acts as multiplication by $b$ on the first or if you prefer to write things multiplicatively like you did above, acting by raising the first generator to the $a$th power) are isomorphic iff the multiplicative orders of $a$ and $b$ in $\mathbb{Z}_p$ are the same.

For instance, $\mathbb{Z}_{101} \rtimes_{19} \mathbb{Z}_{25}$ and $\mathbb{Z}_{101} \rtimes_{36} \mathbb{Z}_{25}$ are not isomorphic (but both make sense since both when raised to the $25$th power are equal to $1$) since $36^5=1 \bmod 101$, but $19^5 \neq 1 \bmod 101$.

Mark O'Brien
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So $\,C_5=\langle x\rangle\;,\;\;C_4=\langle y\rangle\,$ , and suppose the action is inversion

$$x^y:=y^{-1}xy=x^{-1}=x^4$$

Let us check for example

$$a:=(x,y^2) :\;\;a^2=(xx^{y^2},1)=(x^2,1)\;,\;a^3=(x^2,1)(x,y^2)=(x^3,y^2)\;,\;$$

$$a^4=(x^3,y^2)(x,y^2)=(x^4,1)\;,\;\;a^5=(x^4,1)(x,y^2)=(1,y^2)$$

The above automatically means $\,|\langle a\rangle|=10\,$ (why?) and thus we have a cyclic subgroup of order $\,10\,$

DonAntonio
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