Proof check on Cosets: Suppose $G$ is group, $H$ is subgroup, and $g\in G$. Show that $x\in gH \iff x^{-1}\in Hg^{-1}$.

Question

Suppose $G$ is group, $H$ is subgroup, and $g\in G$. Show that $x\in gH \iff x^{-1}\in Hg^{-1}$.

Solution:

Given $G$ is a group, $H <G $ and $g \in G$:

$x\in gH$ then $x=gh$ for some $h \in H$

$x^{-1}=g^{-1}h^{-1}$

$g^{-1}h^{-1}\in Hg^{-1}$.

$x^{-1}\in Hg^{-1}$

Did I do this right and am I missing anything? It seems like I'm missing something or maybe there are some steps that can maybe fill in gaps between some steps I took.

Also you have only proved the forward implication that is $ x \in gH => x^{-1} \in Hg^{-1}$ you need to show the backward as well — tryst with freedom, Mar 21 '22 at 21:23
The double arrow means that when either side is true, the other side is true as well , in other words, when either side is false, both side is false. You have shown that one side being true means the other side is true, now you have to take other side and show that means the initial side you took is true. — tryst with freedom, Mar 21 '22 at 21:24
@Buraian would the steps I took work the other way around. Is every implication a double implication here, where I can replace all my "So" statements with double arrow? — eddie, Mar 21 '22 at 21:26
You would have to check if each sentence is both way true. Also one point is, you don't need to mention $g \in G$ in first line of your solution. That we know already — tryst with freedom, Mar 21 '22 at 21:28
In the solution you have made two mistakes, ending up with the right conclusion. — , Mar 21 '22 at 21:28
@fitzcarraldo I'm not really sure where the mistakes are and how to fix them — eddie, Mar 21 '22 at 21:33
I've edited your question. Usually when writing proof, for neatness and that brevity. Mention where everything comes from in the first line. — tryst with freedom, Mar 21 '22 at 21:37
One more tip is to keep question in block quotes, and to use the 'line' to split between the question text and attempt — tryst with freedom, Mar 21 '22 at 21:42
On line-1 of your solution, it should be $ x \in gH \equiv x= gh$ because it goes both ways. If $x \neq gh $ then $x$ is not in $H$. Final point, each line you should write the reason why that line is true in brackets — tryst with freedom, Mar 21 '22 at 23:45

score 2 · Accepted Answer · answered Mar 21 '22 at 21:35

2

We have

$$\begin{align} x\in gH &\iff x=gh, \text{ some } h\in H\\ &\iff x^{-1}=(gh)^{-1}=h^{-1}g^{-1}, \text{ some } h\in H\\ &\iff x^{-1}\in Hg^{-1}. \end{align}$$

answered Mar 21 '22 at 21:35

Shaun

44,997

You could use there exist instead of 'some' see here, but I suppose it should be written as $( \exists h \in H, x=gh)$ – tryst with freedom Mar 21 '22 at 23:51
1

My use of "some" here, @Buraian, is acceptable. I don't see it used much, to be fair, but a lecturer I had when I was an undergraduate used it all the time. I am aware of the alternatives. Thank you nonetheless. – Shaun Mar 21 '22 at 23:54

Ender_The_Xenocide · Answer 2 · 2022-03-21T21:40:07.040

2

Your proof has some errors and does not include the backward direction. Here is a list of errors:

You should start "Let $g \in G$" instead of doing "for some $g \in G$" later on.
$x^{-1} = h^{-1}g^{-1}$ not $g^{-1}h^{-1}$.
$g^{-1}h^{-1} \not\in Hg^{-1}$.

edited Mar 21 '22 at 21:40

answered Mar 21 '22 at 21:36

Ender_The_Xenocide

390
2
13

1

Thanks for clearing up what I did wrong. This helps a lot! – eddie Mar 21 '22 at 21:38

Proof check on Cosets: Suppose $G$ is group, $H$ is subgroup, and $g\in G$. Show that $x\in gH \iff x^{-1}\in Hg^{-1}$.

Solution:

2 Answers2