Semi direct product-groups that are isomorphic 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.
 A: 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 $19^5 \neq 1 \bmod 101$.  
A: 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 $\,10\,$
