Z6: Identity, Order, Inverse, Generator, Abelian/Non-Abelian, Subgroups

[karush]

$\textit{ For the following groups,}$$(a)\quad \Bbb{Z}_6 \text{ the identity is } \color{red}{0}$
$(b)\quad |\Bbb{Z}_6|=\color{red}{6}$
$(c)\quad |0|=\color{red}{0}$
$(d)\quad |3| =\color{red}{|0,3|}$
$(e)\quad \text{the inverse of 2 is } \color{red}{4}$
$(f)\quad \text{the generator of this group is cyclic groups generated by } \color{red}{ 1}$
$(g)\quad \textit{Abelian/non-Abelian?} \quad \color{red}{Abelian}$
$(h)\quad Z_6 \text{ has $\color{red}{4 }$ subgroups.}$Sorta?

 

[Olinguito]

A couple of mistakes.

$\color{black}(c)\quad |0|=\color{red}1$
$\color{black}(d)\quad \color{red}\langle\color{black}3\color{red}\rangle\color{black}=\color{red}\{0,3\}$

 

[karush]

I tried to check these with W|A but didn't know the input format?

like the next one U(14)

 

[I like Serena]

I tried to check these with W|A but didn't know the input format?

like the next one U(14)

W|A seems to understand:

• 'finite group Z_6'. Note that If we type just 'Z_6' it shows an option to select 'finite group'.
• 'finite group of order 6'. Note that U(14) has order 6. Moreover, it is isomorphic with $\mathbb Z_6$. Isomorphic means that all properties are the same except that the elements have different 'names'.
• 'additive group of integers modulo 6'
• 'multiplicative group of integers modulo 14'

Btw, U(14) is more commonly written as $\mathbb Z_{14}^\times$ or $\mathbb Z_{14}^\ast$. W|A does not seem to understand that either though.

 

[karush]

https://dl.orangedox.com/GXEVNm73NxaGC9F7Cy